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ADVANCED    ALGEBRA 

FOR     <^yi-xD^ 
COLLEGES    AND    SCHOOLS 


BY 

WILLIAM  J.   MILNE,  Ph.D.,  LL.D. 

PRESIDENT   OF  NEW  YORK  STATE  NORMAL   COLLEGE,   ALBANY,    N.Y. 


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NEW  YORK-:.  CINCINNATI-:.  CHICAGO 

AMERICAN    BOOK    COMPANY 


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Copyright,  1901  and  1902,  by 
WILLIAM   J.   MILNE. 

Entbekd  at  Stationers'  Hall,  Londok. 
advanced  algebra. 

B-F      3 


PREFACE 

Many  colleges  and  universities  offer  an  optional  examination 
in  "Advanced  Algebra"  in  addition  to  the  examination  in  "Ele- 
mentary Algebra"'  required  of  all  candidates  for  matriculation. 
The  scientific  and  technical  schools  have  increased  from  time  to 
time  the  amount  of  algebra  that  must  be  mastered  before  students 
may  pass  the  entrance  examinations,  until  their  requirements  at 
present  for  admission  are  substantially  equal  to  what  is  commonly 
included  under  "  Advanced  Algebra."  This  work  is  designed  to 
prepare  students  in  a  thorough  manner  to  meet  both  of  these 
tests. 

The  earlier  pages  of  the  work  are  identical  with  the  author's 
Academic  Algebra,  but  several  of  the  topics  discussed  in  that 
book  have  been  modified  in  treatment  or  enlarged  in  scope,  so 
that  they  may  the  better  meet  the  requirements  of  an  advanced 
course,  and  more  than  one  hundred  sixty  pages  of  new  matter 
have  been  added.  Consequently  this  volume  contains  a  complete 
course  in  both  elementary  and  advanced  algebra. 

The  treatment  is  believed  to  be  sufficiently  full  and  rigorous 
to  meet  the  demands  of  college  courses  in  algebra,  and  to  give  a 
scholarly  basis  for  specializing  in  this  science. 

The  author  desires  especially  to  express  his  indebtedness  to 
Professor  J.  H.  Tanner  of  the  Department  of  Mathematics  at 
Cornell  University  for  valuable  suggestions  regarding  the  treat- 
ment of  several  topics. 

WILLIAM  J.    MILNE. 

Statk  Normal  Colleob, 
Albany,  N.Y. 


28177U 


p 


CONTENTS 


L 


llgebraic  Solutions 
1)kfimtions  and  Notation 
Addition    . 

nSubtraction 

V^Iui-ti  plication 

Division 

v^Review 

>f  actor  ing 

V  Review  of  Factoring 

V 

Highest  CommOx  Divisor 

Lowest  Common  Multiple 
>  Fractions  .... 

Review  of  Fractions     . 
Simple  Equations    . 

Simultaneous  Simple  Equations 

Involution 

Evolution 
y  Theory  ok  J]xponents    . 
'^KRadicals   .... 

Review       .... 

Quadratic  Equations     . 

General  Review 

Ratio  and  Proportion  . 

variation 


6  CONTENTS 

VA6K 

PUOGRESSIONS      .           « 344 

K  Imaginary  and  Complex  Numbers      .        .        .        .        .        .  363 

Inequalities 373 

Variables  and  Limits 379 

Incommensurable  Numbers 388 

Interpretation  of  Results 393 

Indeterminate  Equations 400 

Mathematical  Induction 404 

Binomial  Theorem 408 

^  Logarithms 415 

Permutations  and  Combinations 434 

Probability 445 

Simple  Continued  Fractions 458 

Theory  of  Numbers 470 

Determinants • 479 

CONVERGENCY   OF    SeRIES             .            .^ 496 

Undetermined  Coefficients 509 

)^Exponential  and  Logarithmic  Series 524 

Summation  of  Series 533 

Functions  of  a  Single  Variable 545 

Theory  of  Equation^ 562 


^     ^ 


ADVANCED   ALGEBRA 


oJOic 


ALGEBRAIC    SOLUTIONS 

1.  Problem  1.  A  man  had  400  acres  of  corn  and  oats.  If 
there  were  3  times  as  many  acres  of  corn  as  of  oats,  how  many 
acres  were  there  of  each  ? 

Arithmetical  Solution 
A  certain  number  —  the  number  of  acres  of  oats. 
Then,  3  times  that  number  =  the  number  of  acres  of  corn, 

and  4  times  that  number  =  the  number  of  acres  of  both  ; 

therefore,        4  times  that  number  —  400. 

Hence,  the  number  =  100,  the  number  of  acres  of  oats, 

and  3  times  the  number  =  300,  the  number  of  acres  of  corn. 

Algeuraic  Solution 

Let  X  =  the  number  of  acres  of  oats. 

Then,  3x  =  the  number  of  acres  of  corn, 

and  4  X  =  the  number  of  acres  of  both  ; 

therefore,  4  a;  =  400. 

Hence,  x  =  100,  the  number  of  acres  of  oats, 

and  3  X  =  300,  the  number  of  acres  of  corn. 

2.  An  expression  of  equality  between  two  numbers  or  quan- 
'jities  is  called  an  Equation. 

5  a;  =  30  is  an  equation. 

%  A  question  that  can  be  answered  only  after  a  course  of 
:?3aso'aing  is  called  a  Problem. 

%.   The  process  of  finding  the  result  sought  is  called  the  Solu- 
'Oa  of'  the  problem. 

7 


8  ['AlqebrMc  solutions 

6.  The  expression  in  algebraic  language  of  the  conditions  of  a 
problem  is  called  the  Statement  of  the  problem. 

Solve  algebraically  the  following  problems : 

2.  A  horse  and  saddle  cost  $  50.  If  the  horse  cost  4  times 
as  much  as  the  saddle,  what  was  the  cost  of  each  ? 

3.  A  bicycle  and  suit  cost  $90.  How  much  did  each  cost, 
if  the  bicycle  cost  5  times  as  much  as  the  suit  ? 

4.  Of  240  stamps  that  Harry  and  his  sister  collected,  Harry 
collected  3  times  as  many  as  his  sister.  How  many  did  each 
collect  ? 

5.  If  Mr.  Brown  and  his  son  together  had  f  220,  and  Mr. 
Brown  had  10  times  as  much  as  his  son,  how  much  money  had 
each? 

6.  In  a  room  containing  45  students  there  were  twice  as  many 
girls  as  boys.     How  many  were  there  of  each  ? 

7.  A  had  7  times  as  many  sheep  as  B,  and  both  together  had 
608.     How  many  sheep  had  each  ? 

8.  A  and  B  began  business  with  a  capital  of  $  7500.  If  A 
furnished  half  as  much  capital  as  B,  how  much  did  each  furnish  ? 

Suggestion.  — Let  x  =  the  number  of  dollars  A  furnished. 

9.  A  man  bought  a  cow  and  a  calf  for  $  36,  paying  8  times  as 
much  for  the  cow  as  for  the  calf.     What  was  the  cost  of  each  ? 

10.  James  sold  his  pony  and  a  saddle  for  $  60.  If  the  saddle 
sold  for  1  as  much  as  the  pony,  what  was  the  selling  price  of 
each? 

11.  A  certain  number  added  to  twice  itself  equals  96.  What 
is  the  number  ? 

12.  A  farmer  raised  a  certain  number  of  bushels  of  wheat,  4 
times  as  much  corn,  and  3  times  as  much  barley.  If  there  were 
in  all  4000  bushels  of  grain,  how  many  bushels  of  each  kind  did 
he  raise  ? 

13.  A  boy  bought  a  bat,  a  ball,  and  a  glove  for  $  2.25.  If  the 
bat  cost  twice  as  much  as  the  ball,  and  the  glove  cost  3  times  as 
much  as  the  bat,  what  was  the  cost  of  each  ? 


ALGEBRAIC   SOLUTIONS  9 

14.  In  a  fire  B  lost  twice  as  much  as  A,  and  C  lost  3  times 
as  much  as  A.  If  their  combined  loss  was  $  300,  what  was  the 
loss  of  each  ? 

15.  A  house  and  lot  cost  $3000.  If  the  house  cost  4  times  as 
much  as  the  lot,  what  was  the  cost  of  each  ? 

16.  In  a  business  enterprise  the  joint  capital  of  A,  B,  and  C 
was  $  2100.  If  A's  capital  was  twice  B's,  and  B's  was  twice  C's, 
what  was  the  capital  of  each  ? 

17.  John,  William,  and  George  together  had  120  marbles.  If 
William  had  twice  as  many  as  John,  and  George  had  3  times  as 
many  as  John,  how  many  had  each  ? 

18.  In  an  orchard  of  apple,  pear,  and  cherry  trees,  containing 
1690  trees  in  all,  there  were  4  times  as  many  cherry  trees  as 
pear  trees,  and  twice  as  many  apple  trees  as  cherry  trees.  How 
many  trees  were  there  of  each  kind  ? 

19.  A  number  plus  itself,  plus  twice  itself,  plus  4  times  itself, 
is  equal  to  72.     What  is  the  number  ? 

20.  Charles  is  twice  as  old  as  his  younger  brother,  and  half  as 
old  as  his  older  brother.  If  the  sum  of  the  ages  of  the  three 
brothers  is  28  years,  what  is  the  age  of  each  ? 

21.  A  farmer  had  twice  as  many  sheep  as  horses,  and  twice  as 
many  hogs  as  sheep  and  horses  together.  If  there  were  in  all 
360  animals,  how  many  were  there  of  each  kind  ? 

22.  A  tract  of  land  containing  640  acres  was  divided  into  three 
farms,  such  that  the  first  was  3  times  as  large  as  the  second,  and 
the  third  4  times  as  large  as  the  first.  How  many  acres  did  each 
farm  contain  ? 

23.  Three  boys  divided  160  marbles  among  themselves  so  that 
one  of  them  received  twice  as  many  as  each  of  the  others.  How 
many  did  each  receive  ? 

24.  Divide  30  into  two  parts,  one  of  which  is  14  times  the 
other. 

25.  Divide  18  into  three  parts,  such  that  the  first  is  twice  the 
third,  and  the  second  is  3  times  the  third. 


10  ALGEBRAIC   SOLUTIONS 

26.  Divide  21  into  three  parts,  such  that  the  first  is  twice  the 
second,  and  the  second  is  twice  the  third. 

27.  Divide  36  into  three  parts,  such  that  the  first  is  twice  the 
second,  and  the  third  is  equal  to  twice  the  sum  of  the  first  and 
second. 

28.  Three  newsboys  sold  60  papers.  If  the  first  sold  twice  as 
many  as  the  second,  and  the  third  sold  3  times  as  many  as  the 
second,  how  many  did  each  sell  ? 

29.  Henry  earned  a  certain  number  of  dollars  per  week.  With 
4  weeks'  earnings  he  purchased  a  rifle,  and  with  20  weeks'  earn- 
ings, a  bicycle.  If  both  together  cost  $  72,  how  much  did  he  earn 
per  week  ?     How  much  did  the  rifle  cost  ?  the  bicycle  ? 

30.  A  man  sold  some  ducks  for  50  cents  each,  and  the  same 
number  of  geese  for  75  cents  each.  If  for  all  he  received  $  12.50, 
how  many  of  each  did  he  sell  ? 

31.  John  has  5  times  as  much  money  as  James.  James  has 
24  cents  less  than  John.     How  much  has  each  ? 

32.  A.  man  had  675  sheep  in  three  fields.  If  there  were  twice 
as  many  in  the  first  field  as  in  the  second,  and  twice  as  many  in 
the  third  field  as  in  both  of  the  others,  how  many  sheep  were 
there  in  each  field  ? 

33.  A  man  bequeathed  to  his  daughter  twice  as  much  money 
as  to  his  son,  and  to  his  wife  3  times  as  much  as  to  his  daughter. 
If  all  together  received  f  9000,  how  much  did  each  receive  ? 

34.  A  plumber  and  two  helpers  together  earned  $  7.50  per 
day.  How  much  did  each  earn  per  day,  if  the  plumber  earned  4 
times  as  much  as  each  helper  ? 

35.  What  number  added  to  |  of  itself  equals  20  ? 

Solution 
Let  X  =  the  number 

Then,  x  +  |«  =  20, 

fx  =  20, 

i«  =  4. 
Therefore,  x  =  12,  the  number. 


ALGEBRAIC   SOLUTIONS  11 

36.  If  ^  of  a  number  is  added  to  the  number,  the  sum  is  12. 
What  is  the  number  ? 

37.  If  ^  of  a  number  is  added  to  twice  the  number,  the  sum  is 
36.     What  is  the  number  ? 

38.  The  difference  between  4  times  a  certain  number  and  \  of 
the  number  is  30.     What  is  the  number  ? 

39.  The  difference  between  |  of  a  certain  number  and  f  of  it 
is  16.     What  is  the  number  ? 

40.  After  spending  i  of  my  money  and  losing  \  of  it,  I  had 
$  30.     How  much  had  I  at  first  ? 

41.  The  difference  between  twice  a  certain  number  and  |  of 
it  is  20.     What  is  the  number? 

42.  The  number  150  can  be  divided  into  two  parts,  one  of 
whicli  is  f  of  the  other.     What  are  the  parts  ? 

43.  One  part  of  45  is  f  of  the  other.     What  are  the  parts  ? 

44.  Find  two  parts  of  30  such  that  one  is  ^  of  the  other. 

45.  To  A,  B,  and  C  I  owe  in  all  $93.  If  I  owe  A  |  as  much 
as  C,  and  B  f  as  much  as  C,  how  much  do  I  owe  each  ? 

46.  The  length  of  a  field  is  If  times  its  width,  and  the  distance 
around  the  field  is  120  rods.  If  the  field  is  rectangular,  what  are 
its  dimensions  ? 

47.  A,  B,  C,  and  D  buy  $  16,000  worth  of  railroad  stock.  How 
much  does  A  take,  if  B  takes  3  times  as  much  as  A,  C  twice 
as  much  as  A  and  B  together,  and  D  ^  as  much  as  A,  B,  and  C 
together  ? 

48.  In  one  season  an  orchard  bore  650  bushels  of  fruit,  con- 
sisting of  I  as  many  bushels  of  pears  as  of  peaches,  and  twice 
as  many  bushels  of  apples  as  of  pears.  How  many  bushels  were 
there  of  each  ? 

49.  A  horse,  harness,  and  carriage  cost  f  340.  If  the  horse 
cost  3  times  as  much  as  the  harness,  and  the  carriage  cost  1^ 
times  as  much  as  the  horse,  what  was  the  cost  of  each  ? 


DEFINITIONS   AND    NOTATION 


6.  The  ideas  of  number  and  the  knowledge  of  the  processes 
with  abstract  numbers  that  the  student  has  gained  in  arithmetic 
are  a  proper  and  necessary  introduction  to  his  work  in  algebra; 
but  since  number  is  discussed  in  a  more  general  way  in  algebra 
than  in  arithmetic,  many  arithmetical  processes,  terms,  and  sym- 
bols, as  'addition,'  'subtraction,'  'greater,'  'less,'  'exponent,'  '+,' 
'— ,'  etc.,  must  now  be  extended  in  meaning  and  application. 

For  example,  in  an  arithmetical  sense  8  cannot  be  subtracted 
from  5,  nor  does  8»  have  any  meaning ;  but  in  an  algebraic  sense, 
as  will  be  shown  hereafter,  8  can  be  subtracted  from  5  and  8*  is  as 
intelligible  as  8^ 

Indeed,  the  processes  and  principles  of  arithmetic  are  but  spe- 
cial cases  of  the  more  fundamental  processes  and  principles  of 
algebra. 

7.  A  imit  or  an  aggregate  of  units  is  called  a  Whole  Number, 
or  an  Integer. 

One  of  the  equal  parts  of  a  unit  or  an  aggregate  of  equal  parts 
of  a  unit  is  called  a  Fractional  Number. 

Such  numbers  are  called  Arithmetical,  or  Absolute  Numbers. 

8.  Arithmetical  numbers  have  fixed  and  known  values,  and  are 
represented  by  symbols  called  numerals;  as  1,  2,  3,  etc.,  Arabic 
figures,  and  I,  V,  X,  etc.,  Roman  letters. 

9.  It  is  often  convenient,  in  solving  a  problem,  to  employ 
letters,  such  as  x,  y,  z,  to  represent  the  numbers  whose  values  are 
sought ;  and,  in  stating  a  rule,  to  employ  letters  to  represent  the 
numbers  that  must  be  given  whenever  the  rule  is  applied. 

Numbers  represented  by  letters  are  called  Literal  Numbers. 

12 


DEFINITIONS  AND  NOTATION  13 

For  example,  the  volume  of  any  rectangular  prism  is  equal  to 

the  area  of  the  base  multiplied  by  the  height.     By  using  v  for 

volume,  a  for  area  of  base,  and  h  for  height,  this  rule  is  abbreviated 

to 

V  =  a  X  h. 

When      a  =  60  and  h  =  5,  v  =  60  x  5  =  300 ; 

when  a  =  36  and  h  =10,  v  =  36  x  10  =  360 ;  etc. 

In  each  problem  to  which  this  rule  applies  a  and  h  represent 
fixed,  known  values,  but  in  consequence  of  being  used  for  all 
problems  of  this  class,  a  and  h  represent  numbers  to  which  any 
arithmetical  values  whatever  may  be  assigned.  Hence,  the  arith- 
metical idea  of  number  is  extended  as  follows. 

10.  A  literal  number  to  which  any  value  can  be  assigned  at 
pleasure  is  called  a  General  Number. 

11.  A  number  whose  value  is  known  or  a  number  to  which 
any  value  can  be  assigned  is  called  a  Known  Number. 

The  numerals,  3  and  4|,  and  the  general  numbers  a  and  ft  in  w  =  a  x  h,  in 
§  9,  are  known  numbers. 

Known  literal  numbers  are  generally  represented  by  the  first 
letters  of  the  alphabet. 

12.  A  number  whose  value  is  to  be  found  is  called  an  Unknown 
Number. 

Unknown  numbers  are  usually  represented  by  the  last  letters  of 
the  alphabet. 

ALGEBRAIC  SIGNS 

13.  The  Sign  of  Addition  is  +,  read  ^  plus  J 

It  indicates  that  the  number  following  it  is  to  be  added  to  the 
number  preceding  it. 

a  -I-  6,  read  '  a  plus  6,'  indicates  that  b  is  to  be  added  to  a. 

14.  The  Sign  of  Subtraction  is  — ,  read  '  minus.'' 

It  indicates  that  the  number  following  it  is  to  be  subtracted 
from  the  number  preceding  it. 

a  —  b,  read  '  a  minus  6,'  indicates  that  6  is  to  be  subtracted  from  o. 


14  DEFINITIONS   AND  NOTATION 

15.  The  Sign  of  Multiplication  is  x  or  •,  read  '■  multiplied  hy.^ 

It  indicates  that  the  number  preceding  it  is  to  be  multiplied  by 
the  number  following  it. 

o  X  6,  or  a-h,  indicates  that  a  is  to  be  multiplied  by  h. 

The  sign  of  multiplication  is  usually  omitted  in  algebra,  except 
between  figures. 

a  X  6,  or  a-h,  may  be  abbreviated  to  ab.x  x  y  to  xy,  i  x  bioi  b,  etc.  But 
3x5  cannot  be  written  35,  because  35  means  30  +  5. 

16.  The  Sign  of  Division  is  -r-,  read  'divided  by.' 

It  indicates  that  the  number  preceding  it  is  to  be  divided  by 
the  number  following  it. 

a  ~  b  indicates  that  a  is  to  be  divided  by  b. 

Division  may  be  indicated  also  by  writing  the  dividend  above 
the  divisor  with  a  line  between  them. 

Such  indicated  divisions  are  called  Fractions.     (Cf.  §  158.) 

-  indicates  that  a  is  to  be  divided  by  b. 
b 

17.  The  Sign  of  Equality  is  =,  read  '  is  equal  to'  ov  ' equals.' 

18.  The  Sign  of  Inequality  is  >  or  <. 

When  used  between  two  numbers,  it  signifies  that  they  are 
unequal,  the  greater  number  being  at  the  opening  of  the  sign, 
a  >  6  is  read  '  a  is  greater  than  6.' 
a;  <  5  is  read  '  x  is  less  than  5.' 

-  19.  The  Signs  of  Aggregation  are :  the  Parenthesis,  (  ) ;  the 
Vincxdum,  — ;  the  Brackets,  [  ]  ;  the  Braces,  \  \  ;  and  the  Vertical 
Bar,  \. 

They  show  that  the  expressions  included  by  them  are  to  be 
treated  as  single  numbers. 

Thus,  each  of  the  forms  (a  +  6)c,  a  -ir  b-c,  [a  +  6]c,  {a  +  6}c,  and  ale, 
signifies  that  the  sum  of  a  and  b  is  to  be  multiplied  by  c.  .  +  b\ 

When  numbers  are  included  by  any  of  the  signs  of  aggregation,  they  are 
commonly  said  to  be  in  parenthesis. 

20.  The  Sign  of  Continuation  is  •  •  •  •  or ,  read  '  and  so  on,' 

or  '  and  so  on  to.' 

2,  4,  6,  8, is  read  '  2,  4,  6,  8,  and  so  on.' 

21.  The  Sign  of  Deduction  is  .-..     It  signifies  therefore  or  hence. 


DEFINITIONS  AND  NOTATION  15 

FACTORS,  POWERS.  AND  ROOTS 

22.  Each  of  two  or  more  numbers  which  multiplied  together 
produce  a  given  number  is  called  a  Factor  of  the  number. 

Since  12  =  2  x  6,  or  4  x  3,  each  of  these  numbers  is  a  factor  of  12. 
Since  3  a&  =  3  x  rt  x  6,  each  of  these  numbers  is  a  factor  of  3  ab. 

23.  When  a  factor  of  a  number  is  considered  as  the  multiplier 
of  the  remaining  factor,  it  is  called  a  Coefficient  of  that  factor. 

In  7  X,  5  ax,  bxy,  and  (a  —  b)x,  the  coefficients  of  x  are  7,  5  a,  by,  and 
(a  —  6) ;  in  bxy,  bx  is  the  coefficient  of  y. 

Coefficients  are  Numerical,  Literal,  or  Mixed,  according  as  they 
are  composed  oificjures,  letters,  or  hoth.  figures  and  letters. 

When  no  numerical  coefficient  is  expressed,  the  coefficient  may 
be  considered  to  be  1. 

24.  When  a  number  is  used  a  certain  number  of  times  as  a 
factor,  the  product  is  called  a  Power  of  the  number. 

Powers  are  named  from  the  number  of  times  the  number  is  used 
as  a  factor. 

When  a  is  used  twice  as  a  factor,  the  product  is  the  second  power  of  a  ; 
when  a  is  used  three  times  as  a  factor,  the  product  is  the  third  power  of  a  ; 
four  times,  the  fourth  power  of  a  ;  n  times,  that  is,  any  number  of  times,  the 
nth  power  of  a. 

Tlie  second  power  is  also  called  the  square,  and  the  third  power  the  cube. 

The  product  indicated  by  a  x  a  x  a  x  a  X  a  may  be  more  briefly 
indicated  by  a^.  Likewise,  if  a  is  to  be  used  n  times  as  a  factor, 
the  product  may  be  indicated  by  a". 

25.  A  figure  or  letter  placed  a  little  above  and  to  the  right  of 
a  number  is  called  an  Index  or  an  Exponent  of  the  power  thus 
indicated. 

The  integers  that  the  student  has  been  using  in  arithmetic  have  been 
positive  integers. 

When  the  exponent  is  a  positive  integer,  it  indicates  the  num- 
ber of  times  that  the  number  is  to  be  used  as  a  factor. 

52  indicates  that  5  is  to  be  used  twice  as  a  factor ;  a*  indicates  that  a  is  to 
be  used  3  times  as  a  factor. 

When  no  exponent  is  written,  the  exponent  is  regarded  as  1. 
5  is  regarded  as  the  first  power  of  5,  and  a^  is  usually  written  a. 


16  DEFINITIONS  AND  NOTATION 

The  terms  coefficient  and  exponent  should  be  carefully  distin- 
guished. 

Thus,     5a  =  a  +  a  +  a  +  a  +  rt,  but  a^^axaxaxaxa. 

26.  One  of  the  equal  factors  of  a  number  is  called  a  Root  of  the 
number. 

5  is  a  root  of  25  ;  a  is  a  root  of  a*  ;  4  a!  is  a  root  of  64  y?. 

Koots  are  named  from  the  number  of  equal  factors  into  which 
the  number  is  separated. 

One  of  the  tioo  equal  factors  of  a  number  is  its  second  root ;  one  of  the 
three  equal  factors  of  a  number  is  its  third  root ;  one  of  the  four  equal  factors, 
the  fourth  root;  one  of  the  n  equal  factors,  the  nth  root. 

The  second  root  of  a  number  is  also  called  its  square  root,  and  its  third 
root  is  called  its  cube  root. 

27.  The  symbol  which  denotes  that  a  root  of  a  number  is  sought 
is  y',  written  before  the  number. 

It  is  called  the  Root  Sign,  or  the  Radical  Sign. 
A  figure  or  letter  written  in  the  opening  of  the  radical  sign 
indicates  what  root  of  the  number  is  sought. 
It  is  called  the  Index  of  the  root. 
When  no  index  is  written,  the  second,  or  square  root  is  meant. 

v^  indicates  that  the  third,  or  cube  root  of  8  is  sought. 

Vox  and  Va  —  b  indicate  the  square  roots  of  ax  and  a  —  b,  respectively. 

The  horizontal  line  used  in  connection  with  the  radical  sign  is  a 
vinculum. 

ALGEBRAIC  EXPRESSIONS 

28.  A  number  expressed  by  algebraic  symbols  is  called  an 
Algebraic  Expression. 

29.  When  signs  of  operation  are  employed  in  algebraic  ex- 
pressions, the  sequence  of  operations  is  determined  by  the  follow- 
ing conventional  law : 

A  series  of  additions  and  subtractions  or  of  multiplications  and 
divisions  are  performed  in  order  from  left  to  right. 
3-1-4-2-1-3=    7-2-1-3  =  6-1-3=    8. 
3x4H-2x3  =  12-f-2x3  =  6x3  =  18. 

a  -\-b  —  c  +  d  indicates  that  6  is  to  be  added  to  a,  then  from  this  result  c 
is  to  be  subtracted,  and  to  the  result  just  obtained  d  is  to  be  added. 


DEFINITIONS  AND  NOTATION  17 

30.   When  a  particular  number  takes  the  place  of  a  letter  or 
general  number,  the  process  is  called  Substitution. 

KuMERicAL  Substitutions 

1.    When  a  =  2,  ?>  =  3,  and  c  =  5,  what   are    the    numerical 
values  of  3  c,  (?,  V8  ab^,  or  +  h-,  and  (a  +  6)^,  respectively  ? 

Solutions 
.7c  =  3-5=:15. 

c8  =  5  •  5  •  5  =  125. 


VS"^  =  V8  •  2  .  3  .  3  =  V2  ■  2  X  2  .  2  X  3  .  3  =  2  x  2  X  3  =  12. 

a2  +  6'^  =  2  •  2  +  3  .  3  =  4  +  9  =  13.' 

(a  +  6)2  =(a  +  6)(a  +  6)  =  (2  +  3)(2  +  3)=  5  •  5  =  25. 

Find  the  numerical  value  of  each  of   the  following  algebraic 
expressions,  when  a  =  5,  6  =  3,  c  =  10,  m  =  4,  ?t  =  1 : 

2.  10  a.  11.    (ahy.  19.    .^/T^i^. 

3.  2  ah.  12.    a^ft^. 

20     ^  +  ^^ 

4.  3  cm.  ,„        /7> ,.      9^ 

13.    V2  acn.  c  —  zm 

^'        ^'  14.   3b^cn\  oi         .       *?-« 

6.  ocml  21.    c  + 

7.  2  a^o. 


f  —  2  ?;i 
22.    a'^c. 


8.  3  6ml  16.    (a-?.)l 

9.  3a«6.  17.    {n  +  iy.  23.    m"  ». 
10.    am*.                      18.    w'^  +  1.                      24.    (bm)"-^. 

31.  An  algebraic  expression  whose  parts  are  not  separated  by 

+  or  —  is  called  a  Term ;  as  2a^,  —  5  xyz,  and  —  • 

z 

In  the  expression  2  a;^  —  5  xyz  +  ?^  there  are  three  terms. 

z 
The  expression  rtt(a  +  6)  is  a  term,  the  parts  being  in  and  (a  +  6). 

32.  Terms  that  contain  the  same  letters  with  the  same  expo- 
nents are  called  Similar  Terms. 

3  x^  and  12  x^  are  similar  terms  ;  also  3(a  +  6)2  and  12(a  +  6)2  ;  also  ax 
and  6x,  regarding  a  and  6  as  the  coefiBcients  of  x. 

ALG.  — 2 


18  DEFINITIONS   AND   NOTATION 

33.  Terms  that  contain  different  letters,  or  tlie  same  letters 
with  different  exponents,  are  called  Dissimilar  Terms. 

5  a  aiKl  3  by  are  dissimilar  terms ;  also  8  d^b  and  3  ab'^. 

34.  Each  literal  factor  of  a  term  is  called  a  Dimension  of  the 
term. 

The  number  of  literal  factors  or  dimensions  of  a  term  indicates 
its  Degree. 

abed  is  a  term  of  the  fourth  degree,  because  it  is  composed  of  four  literal 
factors,  or  has  four  dimensions,  x^  is  a  term  of  the  third  degree,  since  x^  is 
a  convenient  way  of  indicating  that  x  is  taken  three  times  as  a  factor.  The 
expressions  4x:^y2^  and  ^xyz*  are  each  of  the  sixth  degree. 

35.  The  term  of  highest  degree  in  an  expression  determines  the 
Degree  of  the  Expression. 

x^  +  3  x-  +  X  +  2  and  abc  +  b-c  +  ac  —  b  are  expressions  of  the  third  degree. 

36.  When  all  the  terms  of  an  expression  are  of  the  same  degree, 
the  expression  is  called  a  Homogeneous  Expression. 

afi  +  3 x^y  +  xy^ -\-2  y^  and  abc  +  b-c  +  ac-  are  each  homogeneous  expressions. 

37.  An  algebraic  expression  of  one  term  only  is  called  a 
Monomial,  or  a  Simple  Expression. 

xy  and  3  ab  are  monomials. 

38.  An  algebraic  expression  of  more  than  one  term  is  called  a 
Polynomial,  or  a  Compound  Expression. 

3  a  +  2  6,  xy  +  yz  +  zx,  and  a^  +  6^  —  c^  +  2  ab  are  polynomials. 

39.  A  polynomial  of  two  terms  is  called  a  Binomial. 
Sa  +  2b  and  x^  —  y'^  are  binomials. 

40.  A  polynomial  of  three  terms  is  called  a  Trinomial. 

a  +  b  +  c  and  3x  —  2y—z  are  trinomials. 

41.  An  expression,  any  term  of  which  is  a  fraction,  is  called  a 
Fractional  Expression. 

*"  —  3  X  +  -  is  a  fractional  expression. 
a^  X 

42.  An  expression  that  contains  no  fraction  is  called  an  Inte- 
gral Expression. 


DEFINITIONS  AND  NOTATION  19 

5  a2  _  2  a  and  6  x  are  integral  expressions. 

Expressions  like  x^  -\-\x'^ +  \x+\  are  sometimes  regarded  as  integral, 
since  the  literal  numbers  are  not  in  fractional  form. 

43.  An  expression  that  can  be  written  without  using  a  root 
sign  is  called  a  Rational  Expression. 

1,  2,  3,  •••,  a  +  i.  — — ,  and  (a  -  6)^  are  rational  expressions. 

'  x-y 
V25  is  rational,  since  it  can  be  written  5  without  a  root  sign. 

44.  An  expression   that   cannot   be  written  without   using   a 
root  sign  is  called  an  Irrational  Expression. 

a  +  Vh,  a  +  2  Va  +  1,  and  v^4  are  irrational  expressions. 
Va^  is  not  irrational,  however,  since  it  may  be  written  a. 


POSITIVE   AND    NEGATIVE   NUMBERS 

46.    For  convenience,  arithmetical  numbers  may  be  arranged 
in  an  ascending  scale : 

0,      1,      2,      3,      4,      5,       ... 

1 1 I I \ I 


The  operations  of  addition  and  subtraction  are  thus  reduced  to 
counting  along  a  scale  of  numbers.  2  is  added  to  3  by  beginning 
at  3  in  the  scale  and  counting  2  units  in  the  ascending,  or  additive 
direction;  and  consequently,  2  is  subtracted  from  3  by  beginning 
at  3  and  counting  2  units  in  the  descending,  or  subtractive 
direction.  In  the  same  way  3  is  subtracted  from  3.  But  if  we 
attempt  to  subtract  4  from  3,  we  discover  that  the  operation  of 
subtraction  is  restricted  in  arithmetic,  inasmuch  as  a  greater 
number  cannot  be  subtracted  from  a  less.  If  this  restriction 
held  in  algebra,  it  would  be  impossible  to  subtract  one  literal 
number  from  another  without  taking  into  account  their  arith- 
metical values.  Therefore,  this  restriction  must  be  removed  in 
order  to  proceed  with  the  discussion  of  numbers. 

To  subtract  4  from  3  we  begin  at  3  and  count  4  units  in  the 
descending  direction,  arriving  at  1  on  the  opposite,  or  subtractive 
side  of  0.  It  now  becomes  necessary  to  extend  the  scale  1  unit 
in  the  subtractive  direction  from  0. 


20  DEFINITIONS  AND  NOTATION 

To  subtract  5  from  3  we  begin  at  3  and  count  5  units  in  the 
descending  direction,  arriving  at  2  on  the  opposite,  or  subtractive 
side  of  0.  The  scale  is  again  extended,  and  may  be  extended 
indefinitely  in  the  subtractive  direction  in  a  similar  way. 

For  convenience,  numbers  on  opposite  sides  of  0  are  dis- 
tinguished by  means  of  the  small  signs  "*"  and  ~,  called  signs  of 
quality,  or  direction  signs,  +  being  prefixed  to  those  which  stand 
in  the  additive  direction  from  0  and  ~  to  those  which  stand  in 
the  subtractive  direction  from  0. 

The  former  are  called  Positive  Numbers,  the  latter  Negative 
Numbers. 

Hence,  the  scale  of  algebraic  numbers  may  be  written : 

...,      -5,     -4,    -3,     -2,     -1,      0,    +1,    +2,    +3,    +4,     +5,        ... 

I  I  I \ ! \ I I \ 1  I 

46.  By  repeating  +1  as  a  unit  any  positive  number  may  be 
obtained,  and  by  repeating  ~1  as  a  unit  any  negative  number 
may  be  obtained.  Hence,  positive  numbers  are  measured  by 
the  positive  unit,  "^1,  and  negative  numbers  by  the  negative  unit,  ~1, 
or  by  parts  of  these  units. 

47.  If  +1  and  ~1,  or  +2  and  ~2,  or  any  two  numbers  numerically 
equal  but  opposite  in  quality  are  taken  together,  they  cancel 
each  other.  For  counting  any  number  of  units  from  0  in  either 
direction  and  then  counting  an  equal  number  of  units  from  the 
result  in  the  opposite  direction,  we  arrive  at  0.     Hence, 

If  a  positive  and  a  negative  number  are  united  into  one  mimher, 
any  number  of  units  or  parts  of  units  of  one  cancels  an  equal  number 
of  units  or  parts  of  units  of  the  other. 

48.  Two  concrete  quantities  of  the  same  kind  are  sometimes 
opposed  to  each  other  in  some  sense  so  that,  if  united,  any 
number  of  units  of  one  cancels  an  equal  number  of  units  of  the 
other.  For  convenience,  such  quantities  are  often  distinguished 
as  positive  and  negative. 

If  money  gained  is  positive,  money  lost  is  negative,  for  any  sum  gained 
is  canceled  by  an  equal  sum  lost.  If  a  rise  in  temperature  is  positive,  a  fall 
in  temperature  is  negative.  If  distances  )iorth  or  west  or  upstream  are 
positive,  distances  south  or  east  or  downstream  are  negative. 


ADDITION 


49.  1.  If  a  man  has  10  dollars  in  one  pocket  and  15  dollars  in 
another,  how  much  money  has  he  ? 

2.  If  in  algebra  money  in  hand  is  considered  a  positive  quan- 
tity, indicate  his  iinancial  condition  algebraically.  What  is  the 
sum  of  10  positive  units  and  15  positive  units,  that  is,  of  +10  and 
+15  ?  of  +4  and  +8  ?  of  +a  and  +6  ? 

3.  If  a  person  owes  one  man  10  dollars  and  another  15  dollars, 
how  much  does  he  owe  both  ?  Indicate  his  financial  condition 
algebraically,  regarding  a  debt  as  a  negative  quantity. 

4.  What  is  the  sum  of  10  negative  units  and  15  negative  units, 
that  is,  of  -10  and  "15  ?   of  "6  and  "14  ?   of  "a  and  'h  ? 

5.  What  sign  has  the  sum  of  two  algebraic  numbers  that  have 
like  signs  ? 

6.  If  a  man  has  25  dollars  and  owes  15  dollars,  how  much  of 
his  money  will  be  required  to  cancel  the  debt?  How  many 
dollars  will  he  have  after  settlement  ? 

7.  What  is  the  result  when  "15  is  united  with  +25,  that  is, 
what  is  the  algebraic  sum  of  ~15  and  +25  ?  of  ~20  and  ^10  ?  of  +8 
and -3?   of +6  and  "10? 

50.  The  aggregate  value  of  two  or  more  algebraic  numbers  is 
called  their  Algebraic  Sum. 

The  process  of  finding  the  simplest  expression  for  the  algebraic 
sum  of  two  or  more  numbers  is  called  Addition. 

51.  Principles.  —  1.  The  algebraic  sum  of  two  numbers  with 
like  signs  is  equal  to  the  sum  of  their  absolute  values  with  the  com- 
mon sign  prefixed. 

21 


22  ADDITION 

2.  Tlie  algebraic  sum  of  two  numbers  with  unlike  signs  is  equal 
to  the  difference  between  their  absolute  values  with  the  sign  of  the 
numerically  greater  prefixed. 

By  successive  applications  of  the  above  principles  any  number 
of  numbers  may  be  added. 

Only  similar  terms  can  be  united  into  a  single  term. 

Principle  1  may  be  established  as  follows : 

The  sum  of  5  positive  units  and  3  positive  units  is  evidently  (5  +  3)  posi- 
tive units,  or  8  positive  units  ;  tliat  is, 

+5 +  +3  =  +(5 +  3)  =+8. 

Similai-ly,  whatever  absoUite  values  a  and  h  represent,  since  a  times  the 
unit  +1  plus  6  times  the  unit  +1  is  equal  to  (a  +  6)  times  the  unit  +1, 
§  46,  +a  +  +6  =  +(o  +  h). 

Again,  the  sum  of  5  negative  units  and  3  negative  units  is  (5  +  3)  negative 
units,  or  8  negative  units  ;  that  is, 

-5  +  -3  =  -(5  +  3)  =  -8. 

Similarly,  whatever  absolute  values  a  and  h  represent,  since  a  times  the 
unit  -1  plus  h  times  the  unit  -1  is  equal  to  (a  +  b)  times  the  unit  ^1, 
§  46,  -a  +  -6  =  -{a  +  h). 

Principle  2  may  be  established  as  follows : 

The  sum  of  5  positive  units  and  3  negative  units  is  2  positive  units,  since, 
§  47,  the  3  negative  units  cancel  3  of  the  positive  units  and  leave  2  positive 

units ;  that  is, 

+5  + -3  =  +(5 -3)  =  +2. 

The  sum  of  5  positive  units  and  7  negative  units  is  2  negative  units,  since, 

§  47,  the  5  positive  units  cancel  5  of  the  negative  units  and  leave  2  negative 

units ;  that  is, 

+5 +  -7  =-(7 -5)  =-2. 

Similarly,  whatever  absolute  values  a  and  h  represent, 
if  a>6,  +a  + -6  = +(a -6), 

for,  §  47,  the  h  negative  units  will  cancel  b  of  the  a  positive  units  and  leave 
(a  —  b)  positive  units  ; 

but  if  ft  >  a,  +a  +  -b  =  -(b  -  a), 

for,  §  47,  the  a  positive  units  will  cancel  a  of  the  b  negative  units  and  leave 
(6  —  a)  negative  units. 

Examples 

-3  +  +4  ^_  -2  +  +8  +  -9. 
+m  +  ~n,  if  m>n. 


Find  the  value  of 

1.    +7  4-  +3.            3. 

+7  +  -3. 

5. 

2.    -7 +  -3.            4. 

-7  +  +3. 

6. 

ADDITION  23 

52.  To  conform  with  the  ideas  already  presented,  the  terms 
<  greater '  and  '  less '  must  be  interpreted  as  follows  : 

An  algebraic  number  is  increased,  or  made  greater,  when  a  posi- 
tive number  is  added  to  it,  and  decreased,  or  made  less,  when 
a  negative  number  is  added  to  it. 

Since,  by  §  51,  -3  +  ^1  ="2,  -2  + +1  = -f,  -f  + +1  =  0, 
+1  -|-  +1  =  +2,  etc.,  in  the  scale  of  algebraic  numbers 

...,  -5,  -4,  -3,  -2,  -1,  0,  +1,  +2,  +3,  +4,  +5,  ••., 

each  number  is  greater  than  the  number  on  its  left  and  less  than 
the  number  on  its  right ;  that  is, 

...,  -3<-2,  -2<-l,  -1<0,  0<+l,  +l<+2,  +2<+3,  .... 

Note.     -3  <  2  may  be  read  '  -3  is  less  than  -2 '  or  '  -2  is  greater  than  -3.' 

Hence,  it  follows  that : 

1.  Any  positive  number  is  greater  than  zero  and  any  negative 
number  is  less  than  zero. 

2.  Of  two  positive  numbers  that  ivhich  has  the  greater  absolute 
value  is  the  greater,  and  of  two  negative  numbers  that  which  has  the 
less  absolute  value  is  the  greater. 

63.    Abbreviated  notation  for  addition. 

Referring  to  the  scale  of  algebraic  numbers,  it  is  evident  that 
adding  positive  imits  to  any  number  is  equivalent  to  counting 
them  in  the  positive  direction  from  that  number,  and  adding 
negative  units  to  any  number  is  equivalent  to  counting  them  in 
the  negative  direction  from  that  number.  Hence,  in  addition, 
the  signs  -|-  and  —  denoting  quality  have  primarily  the  same 
meanings  as  the  signs  +  and  —  denoting  arithmetical  addition 
and  subtraction.  For  example,  by  the  definition  of  positive  and 
negative  numbers, 

+1  means  0  +  1  and  "1  means  0  —  1 ; 
also  """5  means  0+5  and  ~5  means  0  —  5 ;  etc. 

Hence,  in  addition,  but  one  set  of  signs,  +  and  — ,  is  necessary, 
and  in  finding  the  sum  of  any  given  numbers,  the  signs  +  and  — 
may  be  regarded  either  as  signs  of  quality  or  as  signs  of  opera- 
tion, though  it  is  commonly  preferable  to  regard  them  as  signs 
of  operation. 


24  ADDITION 

For  brevity,  it  is  customary  to  omit  the  sign  +  before  a 
monomial  or  before  the  first  term  of  a  polynomial.  But  the 
sign  —  cannot  be  omitted. 

+5  +  +3  +  -6  is  written  5  +  3  -  6  ;  -4  +  +8  +  "2  is  written  -4  +  8-2. 
When  there  is  need  of  distinguishing  between  the  signs  of 
quality  +  and  —  and  the  signs  of  operation  +  and  — ,  the  num- 
bers and  their  signs  of  quality  may  be  inclosed  in  parentheses. 

Thus,  if  a  =  +  5,  6  =  -  3,  and  c  =  -  2,  then  a  +  6  +  c=(+5)  +  (-3) 
+  (-2);  a-  &-c  =  (+5)-(-3)-(-2);  a6c  =  (+ 5)(- 3) (- 2);  etc. 

54.  A  term  preceded  by  +,  expressed  or  understood,  is  called 
a  Positive  Term,  and  a  term  preceded  by  — ,  a  Negative  Term. 

Thus,  in  the  polynomial  3a  +  26  —  5c  the  first  and  second  terms  are 
positive  and  the  third  term  is  negative. 

Examples 
Write  the  following  with  one  set  of  signs : 

1.  +7  +  +8.  4.    +10  +  -2  +    --4.  7.    +a  -f  "6. 

2.  +6 +  -5.  5.      -6 +  -3 +  +16.  8.    -a  +  +&  +  -c. 

3.  -3  +  -7.  6.      +8  +  +4  +    -5.  9.     -x  +  -y-[-  -%. 

55.  1.  How  does  5  +  3  —  2  compare  in  value  with  5  —  2  +  3, 
or  with  3-2  +  5,  or  with  -  2  +  3  +  5  ? 

2.  How  does  a-\-h  —  c  compare  in  value  with  h  —  c  -{-a,  or 
with  h  -\-  a  —  c,  or  with  a  —  c-^-h'i 

3.  In  what  order  may  numbers  be  added  ? 

Law  of  Order,  or  Commutative  Law  for  Addition.  —  Algebraic  num- 
bers may  be  added  in  any  order. 

The  Law  of  Order  may  be  established  as  follows : 

We  know  from  arithmetic  that  arithmetical  numbers  may  be  added  in  any 
order.  Since,  §  51,  Prin.  1,  algebraic  numbers  having  like  signs  are  added 
by  prefixing  their  common  sign  to  the  sum  of  their  absolute,  or  arithmetical 
values,  and  since  in  finding  this  sum  the  absolute  values  may  be  added  in 
any  order,  it  follows  that  algebraic  numbers  having  like  signs  may  be  added 
in  any  order. 

If  some  of  the  numbers  are  positive  and  some  are  negative,  §  47,  the  same 
number  of  positive  and  negative  units  will  cancel  each  other,  and  the  same 
number  of  one  or  the  other  will  be  left,  in  whatever  order  the  numbers  are 
added. 


ADDITION  25 

Hence,  whether  the  numbers  have  like  or  unlike  signs,  they  may  be  added 
in  any  order  ;  that  is, 

a  +  6  +  c  =  &  +  c  +  a  =  c  +  rt  +  6  =  c+6  +  a,  etc. , 
for  all  values  of  a,  6,  and  c. 

56.  1.  How  are  the  numbers  4,  \,  2,  and  f  grouped  in  adding? 
the  numbers  25  and  32,  or  20  +  5  and  30  +  2  ? 

2.  In  what  manner  may  the  terms  of  an  expression  be  grouped 
in  addition  ? 

Law  of  Grouping,  or  Associative  Law  for  Addition.  —  TJie  s%im  of 
three  or  more  algebraic  numbers  is  the  same  in  whatever  manner  the 
numbers  are  grouped. 

The  Law  of  Grouping  may  be  established  as  follows : 

By  the  Law  of  Order,       a  +  6  +  c  =  6  +  c  +  a 
§29,  '       =(h  +  c)+a 

by  the  Law  of  Order,  —  a  +{b  -\-  c). 

Other  cases,  as  a  +  6  +  c  =  (a  +  c)  +'6,  etc.,  may  be  proved  similarly. 

Hence,  for  all  values  of  a,  6,  and  c, 

rt4-6  +  c  =  rt4-(&  +  c)  =  (a4-c)+&  =  c  +  (a  +  6),  etc. 

57.  To  add  similar  monomials. 

Examples 

1.  Add  4  a  and  3  a. 

PROCESS         Explanation. — Just  as  4  =  1  +  1  +  1  +  1,  so  4a=a+a  +  a+a; 
4  d         just  as3  =  l  +  l  +  l,  so  3  a  =  a  +  a  + a.     Therefore,  4  a  +  3  a 
=  a  +  a-\-a-\-a-\-a-\-a  +  a,  the  symbol  for  which  is  7  a. 
Or,  the  sum  may  be  obtained  by  adding  the  numerical  coeffi- 
7  a         cients  and  annexing  to  their  sum  the  common  literal  part. 

2.  Add  Aa,  ^a,   —  3  a,  and  i  a. 

PROCESS 

4a+-|a  —  3a  +  ia  =  4a  —  3a  +  (|a4-|a)  =  a+-2a  =  3(x. 

Explanation.  —  By  the  Law  of  Grouping  the  sum  of  f  a  and  ^  a  may 
be  added  to  the  sum  of  4  a  and  —  3  a.  Just  as  4  =  1  +  1  +  1  +  1,  so 
4a  =  rt  +  a  +  a  +  a;  just  as  —  3=— 1  —  1  —  1,  so  — 3a=— a  —  a  —  a. 
Therefore,  ia  —  3a  =  a  +  a  +  a+a  —  a  —  a  —  a=  (by  the  Law  of  Order) 
a  —  a  +  a  —  a  +  a  —  a  +  a  —  0  +  a  =  a.  Just  as  |  =  i  +  i  +  i,  so  |  a 
=  ^a  +  ^a  +  ^a.  Therefore,  |  a  +  |  a  =  J  a  +  ^  a  +  |  a  +  ^  a  =  |  a  =  2  a. 
Adding  the  two  groups,  rt  +  2a  =  a  +  a  +  a  =  3a. 

Or,  the  sum  may  be  obtained  by  adding  in  any  order  or  manner  the 
numerical  coefficients,  and  annexing  to  their  sum  the  common  literal  part. 


3a 


26  ADDITION 

Simplify  the  following : 

3.  2y  -1  y  —  5y  ~y  +  10y  —  &y  +  ?>y. 

4.  5a  — 3a  +  8a  — 10a  — 5a  — lla  +  24a. 

5.  3by -5by- 10 by -Uby-^iSby. 

6.  8  a^b  +  6  a'b  -  11  a^b  -  2  aJ'b  +  9  a%. 

7.  l|ary-ia^2/==-lJi^a^/  +  3iar'/  +  r'/. 

8.  5  (xy)2  -  3  (a;^)2  -  15  (xyf  +  4  (xyf  +  13  (a^)^. 

9.  (a  —  x)  +  5  (a  —  x)  +  7  (a  —  a;)  —  3  (a  —  x)  —  2  (a  —  a;). 

10.  3(a  ~\-by-\-6{a+  by-10(a  +  bf  -  (a  +  bf+  12(a  +  bf. 

11.  20  Vx  -  3  -  8  Va;  -  3  -  12  Vx  -  3  +  Va;  -  3  +  7  Va;  -  3. 

12.  3  a;(af'  -  2  .x-  +  3)  -  a;(a;2  -  2  .t  +  3)  +  2  a;(ar^  -  2  a;  +  3). 

13.  2(a:  -  1)  -  13(a;  -  1)  +  5(a;  _  1)  +  10(a;  -  1)  +  6(a;  -  1). 

14.  i-(a  +  &  -  c)  -  |(a  +  6  -  c)  +  i^(a  +  b  -  c). 

Since  only  similar  terms  can  be  united  into  a  single  term,  in 
algebra  dissimilar  terms  are  considered  to  have  been  added  when 
they  have  been  w^ritten  in  succession  with  their  proper  signs. 

In  algebra  many  indicated  operations  are  regarded  as  per- 
formed. 

Since  5  a,  —  3  b,  and  2  c  cannot  be  united  into  a  single  term,  their  sum 
is  written  5  a  —  3  fe  +  2  c. 

15.  Add  6a,   —5b,  —3a,  3b,  2c,  and  —a. 
Solution.     6a-5b-3a  +  Sb  +  2c-a  =  2a-2b  +  2c. 

Add  the  following: 

16.  2xy,  Aab,  3xy,  and  ab. 

17.  ran,   —3cd,   —6mn,  and  Acd. 

18.  a,   —  b,  2  c,  —2  a,  3  b,  and  —4  c. 

19.  6x,  3y,   —2x,y,  —3x,z,  and  —3y. 

20.  2a,  2  6,  2c,  2c?,  —a,   —3  6,   —  c,  and  —3d 

21.  a,  —4  a,  2  6,  cd,  —  2a6,  5  6,  and  —  3cd. 


ADDITION  27 

58.   To  add  polynomials. 

Examples 

1.  Add  3a-36  +  5c,  — 3a  +  2  6,  and  c  -  4  6  +  2  a. 

PROCESS  Explanation.  — For  convenience  in  adding,  simi- 

3a  —  36  +  5c  lar  terms  are  written  in  the  same  column. 

—  3  a  +  2  6  '^^^  algebraic  sum  of  the  first  column  is  2  a,  of 

2  a  —  4  6  4-     c  *^^  second  —  5  6,  and  of  the  third  +  6  c  ;  and  these 

^numbers  written  in  succession  express  in  its  simplest 

2a  —  56  +  6c  form  the  sum  sought. 

2.  Simplify   1\  a?h  - 1  ah' -^  2  ac"  +  10  db  - 'i  ac^  +  b  a?h  -  4.  ab^ 
+  5  ac=^  +  6^  +  9  a6=^  -  7  a^d  -  2  6=^  +  2  a62  -  8  a6  -  6  a?b. 

PROCESS 

Ila2&-7a&2  +  2ac2  +  10a6+    W 
+  5a26-4a62-4ac2  -263 

-Ta'^ft  +  Qafe^  +  Sac^ 
-&a?h  +  2ab^  -    8a6 


3a''6  +3ac2+    2a6 


Rule.  —  Arrange  the  terms  so  that  similar  terms  stand  in  the 
same  column. 

Find  the  algebraic  sum  of  each  column,  and  write  the  results  in 
succession  with  their  proper  signs. 

3.  Add  2  a  -  3  6,  2  &  -  3  c,  5  c  -  4  a,  10  a  -  5  6,  and  7  &  -  3  c. 

4.  Add  x  +  y  +  z,  x  —  y  -\-z,  y  —  z  —  x,  z  —  x  —  y,  and  05  —  2!. 

Simplify  the  following  polynomials : 

5.  7a;  — 11 2/4-42  —  72  4-1103  — 4y  +  7 y  —  llz  —  4:X-{-y—x—z. 

6.  a  +  3  6  +  5c-,6a  +  d  +  46-2c-26  +  5a-d-t-a  —  6. 

7 .  Ax' - 3xy  -]- 5y^  +  10 xy  -17  y''  -11  x^ - 5xy -{-12x^  -  2xy. 

8.  2xy  —  5y^-\- a^y^  —  7xy-}-3y^  —  4: o^y^  +  5 xy  +  4 ?/^  +  -s^y^. 

9.  2  ai/  —  3  ac  —  4  ay  +  4  ac  —  6  a?/  +  5  ac  -H 11  ay  —  4  oc  —  ay. 


28 


ADDTTTON 


10.  5  am,  —  3  aW  +  4  —  4  am  +  ahn^  —  2  +  5  -f  a^m^  —6  +  3  am. 

11.  6V^  —  6^xy  +  3Vy  —  4Va;  +  6y/xy  —y/x  —  -Vy  +  3Vy 
—  2V^. 

Add  the  following  polynomials : 

12.  7a-3b  +  5c-10d,2b  +  d-Sc-4:e,  5c-6a  +  2(Z-4e, 
86  —  7a  —  8c—  e,  a  —  5c  +  5d  +  lle,  a  —  6  +  c  +  2d  +  e,  and 
5a  —  46  +  2  c. 

13.  5x  —  3y  —  2z,  4:y--2x-^6z,3a  —  2x  —  4:y,4b  —  2z  —  5y, 
a  —  5b,  5y  —  6x,  Sx  +  2y  —  5a  —  2b,  and  6x  —  y  —  2z  +  4b. 

14.  ?n-|-w  —  V/n.w,  ■\/mn  —  2m—3n,  3m-\-2n,  4n  — Vmn— 3m, 
5  Vwin  —  n,  4  m  —  ?i  —  2  Vmn,  5  w  +  2  m  —  3  Vmn,  and  w  —  6  m. 

15.  2c  — 7d  +  6n,  11m  — 3c  — 5 w,  7/i  — 2d  — 8c,  8d  — 3m 
+  10c,  4d  —  3w  —  8m,  m  —  6n,  and  2m  — 3d. 

16.  4a:3_2a;2-7a;  +  l,  x-3  +  3a^+ 5ic-6,  A.x'-^y? +  2-Qx, 
2ar»-2ic2  +  8a;  +  4,  and  2ar' -  3ic2  -  2a;  +  1. 

17.  a'^  +  5a^6+5a6^  +  6^  a*6  -  2a«  +  a«62  -  2  6^  a%^-3a^b^ 
-Aa*b-  a%  and  2  a^  +  a^6  -  2  a^b^  +  2  a26''  -  3  a6*  +  6^. 

18.  a*-2  a36+3 a^b^  3  ab^-4:  b*-2  aW,  3  a36+4 a*-3  a63+4 b\ 
5a^b  +  7  a'b'  -Aab^-S  ¥,  and  a^  -  6  a36  -  8  a^b\ 

19.  5x«-a^  +  7a;-9,  40^*  -  3a;''  + 6a^  + 12,  x^-5x^-x-l, 
4-ar=-x«,  4x^-10a;2  +  3a;«  +  4,  and  a;«  +  ar^  -  3ar^  -  4  a;- 5. 

20.  3(a  +  6)  +  6(6  +  c),  5(a  +  6)-10(6  +  c),  2(a  +  6)  +  (6  +  c), 
3(6  +  c)  -  (a  +  6),  2(6  +  c)  -  10(a  +  6),  and  3(a  +  6)  -  3(6  +  c). 

21.  a;  +  3  (a  +  1)  —  y,  —  (a  + 1)  —  2  a;  +  4  ?/,  and  3  a;  —  4  (a  +  1). 

22.  a^  -  3  a26c  -  6  a6%  a26  -  6^  -  c^  -  3  a6c,  aW +  b''c  +  b(?, 
5  a'bc  +  4  a62c  +  c^,  6^  -  a'b  -  ab\  a'  +  b'c  +  be',  and  2  a6^c  -  2  60^. 

23.  .12a;3_4a^.2_^^_^2,  .4 a;2  - 4 a;  +  .4  -  a;^^  3|-a;-.6+3ar2+2a;3^ 
and  l-^x  + 1.2x^  +  ^0^. 

24.  aa;  —  f  ax^  —  \  ax^,  f  ax*  —  ^  ax^  —  |^  6a;i/,  1 6a;y  —  f  aa;*  —  ^  ab, 
1 6a;y  —  I  a6  +  ^  ax,  and  2  a6  —  f  aa;  +  |  aa^. 


lab,M 

d 


ADDITION  29 


Exercises 


69.  1.  If  a  boy  has  n  marbles  and  buys  10,  how  many  will  he 
then  have  ?     If  he  gives  away  m  of  these,  how  many  will  be  left  ? 

2.  Mary  has  25  cents.  How  many  cents  will  she  have  after 
spending  10  cents  and  earning  a  cents  ?  If  she  has  c  cents,  spends 
b  cents,  and  earns  a  cents,  how  many  cents  will  she  have  ? 

3.  A  boy  who  has  p  marbles  loses  q  marbles,  and  then  buys 
r  marbles.     How  many  does  he  then  have  ? 

4.  James  is  15  years  old.  In  how  many  years  will  he  be  21 
years  old  ?  In  how  many  years  will  he  be  x  years  old  ?  Harry 
is  y  years  old.  How  many  years  older  is  he  than  James?  In 
how  many  years  will  Harry  be  x  years  old  ? 

5.  Edith  is  14  years  old.  How  old  was  she  4  years  ago?  a 
years  ago  ?     How  old  will  she  be  3  years  hence  ?  h  years  hence  ? 

6.  William  is  x  years  old.  How  old  was  he  a  year  ago  ? 
How  old  will  he  be  in  5  years  ?  in  a  years  ?  After  how  many 
years  will  he  be  21  years  old  ?  m  years  old  ?  How  old  will  he 
be  when  he  is  twice  as  old  as  he  is  now  ? 

7.  In  a  certain  family  there  are  five  children  each  of  whom 
is  2  years  older  than  the  one  next  younger.  If  the  youngest  is 
X  years  old,  what  are  the  ages  of  the  others  ? 

8.  A  woman  sold  some  eggs,  and  with  the  money  bought  8 
pounds  of  sugar  and  5  pounds  of  coffee.  If  the  sugar  cost  a 
cents  a  pound,  and  the  coffee  cost  6  cents  a  pound,  how  much  did 
she  receive  for  the  eggs  ? 

9.  What  two  whole  numbers  are  nearest  to  50?  to  »?  to 
X  -f-  5  ?  If  2/  is  an  even  number,  what  are  the  nearest  even 
numbers  ? 

10.  George  is  a  years  younger  than  Henry,  and  h  years 
younger  than  John.     If  John  is  16  years  old,  how  old  is  Henry  ? 

11.  A  man  paid  two  men,  whom  he  owed,  in  the  following 
manner :  To  the  first  he  gave  an  a-dollar  bill,  and  received 
change  amounting  to  h  dollars;  and  to  the  second  he  gave  a  b- 
dollar  bill,  and  received  change  amounting  to  c  dollars.  How 
much  did  he  owe  both? 


30  ADDITION 

Equations  and  Problems 

60.   1.    Simplify  the   equation   2x  —  3x-\-&x-\-bx  —  x  =  21. 

and  find  the  value  of  x. 

Solution 
2a;-3a;  +  6x  +  5x-x  =  27. 
Uniting  terms,  9  x  =  27. 

Hence,  x  =  3. 

Simplify,  and  find  the  value  of  cc : 

2.  lOx-1  x  +  4.x-Qx  +  llx-20x-\-12x  =  4Q. 

3.  13a;  — 6x  — 4a;  +  7x  +  lla;-16a;4- 15a:  =  20. 

4.  25  .T  -  5  X  -  7  a;  -  2  a;  +  14  x  -  10  »  -  12  X  =  36. 

5.  17a;  +  2x-6x  +  4x-12x-30a;  +  40a;  =  75. 

6.  10a;  +  2x  +  3a;  +  4a;  +  lla;  +  12a;4-18a;  =  60. 

7.  16  a;  — 3x  — 5  a;  — Sx-f  10x  +  15x  — 15x  =  50. 

8.  12a;  +  10a;-20a;  +  16x-3a!-2a;  +  2a;  =  75. 

9.  14a;  — 11  a; +  26  a;  — 35a;  — 4  a; +  7  x  + 4  a;  =  16. 

10.  75  a;  -  37  a;  -  40  a;  +  10  a;  -  8  a;  -  6  .a;  +  9  a;  =  21. 

11.  4  x  +  10  a;  -  60  a;  +  48  a;  +  12  a;  +  5  a;  +  2  a;  =  63. 

12.  7  a;  +  11  a;  —  13  a;  +  15  a;  —  17  a;  —  3  a;  +  5  a;  =  25. 

13.  5  a; -15  a; +  25  a;  — 30  a; +  10  a; +  3  a; +  6  a;  =  56. 

14.  a;  +  2  a;  +  3  a;  +  4  a;  +  5  a;  +  6  a; +  7  a; +  8  a;  =144. 

15.  4  a;  +  12  a:  —  17  a;  —  10  a;  +  15  a;  +  a;  +  15  a;  =  400. 

16.  3  a;  +  10  a;  -  20  .T  -  4  a;  +  12  a;  +  3  a;  +  11  a;  =  300. 

Solve  the  following  problems : 

17.  A  man  bequeathed  $10,000  to  3  sons  and  4  daughters,  so 
that  a  son  received  twice  as  much  as  a  daughter.  What  was  the 
share  of  each  daughter,  and  of  each  son  ? 


I 


I 


ADDITION  31 

18.  John  had  twice  as  many  marbles  as  Henry,  and  \  as  many 
as  Charles.     If  they  had  225  marbles  in  all,  how  many  had  each  ? 

19.  A  had  twice  as  much  money  as  B,  who  had  3  times  as 
much  money  as  C.  If  all  together  had  $  2000,  how  much  money 
had  each? 

20.  A  merchant  owes  A  a  certain  sum  of  money,  B  i  as  much, 
and  C  twice  as  much  as  A.  Vaiious  persons  owe  him  in  all  12 
times  as  much  as  he  owes  B.  If  all  these  debts  were  paid,  he 
would  have  $  10,000.     What  are  the  amounts  he  owes  ? 

21.  Mr.  Jones  succeeded  in  doubling  his  capital  once  every 
5  years.  If  his  capital  at  the  end  of  20  years  was  $  150,000,  with 
what  capital  did  he  begin  ? 

22.  The  distance  around  a  rectangular  field  4  times  as  long  as 
it  is  wide  is  200  rods.     What  are  the  dimensions  of  the  field  ? 

23.  What  are  the  dimensions  of  a  rectangular  field  whose 
length  is  twice  its  width,  if  240  rods  of  fence  are  required  to 
inclose  it  ? 

24.  Two  boys  caught  the  same  number  of  fish,  another  caught 
10  more,  and  another  10  less.  If  they  caught  in  all  120  fish,  how 
many  did  each  catch  ? 

25.  The  sum  of  3  consecutive  whole  numbers  is  84.  What 
are  the  numbers? 

Suggestion.  —  Let  x  represent  the  middle  number.  Then,  the  other  two 
numbers  will  be  represented  by  x  —  1  and  x  +  1. 

26.  Of  what  3  consecutive  even  numbers  is  150  the  sum  ? 

27.  A,  B,  C,  and  D  together  have  $  1500.  If  A  had  f  50  more 
and  B  $  50  less,  they  would  each  have  the  same  sum  as  C  and  i 
as  much  as  D.     How  much  money  has  each  ? 

28.  The  ages  of  4  brothers  differ  successively  by  2  years.  If 
the  sum  of  their  ages  is  56  years,  what  is  the  age  of  each  ? 

29.  Three  newsboys  sold  270  papers  in  an  evening.  If  the 
second  sold  5  less  than  twice  as  many  as  the  first,  and  the  third 
5  more  than  3  times  as  many  as  the  first,  how  many  papers  did 
each  sell? 


SUBTKACTION 


61.  1.  What  is  left  when  5xy  is  taken  from  12  xy?  What  is 
the  sum  of  ~b  xy  and  +12  xy  ? 

2.  What  is  left  when  +3  mn  is  subtracted  from  "•"10  mn  ?  What 
is  the  sum  of  ~3  mn  and  """10  mn  ? 

3.  Instead  of  subtracting  a  positive  number,  what  may  be 
done  to  obtain  the  same  result  ? 

4.  What  is  the  result  when  8  a  is  subtracted  from  10  a  ?  When 
(8a—  5a)  is  subtracted  from  10 a ?  How  does  the  second  result 
compare  with  the  first  ?  What  effect  upon  the  result  has  the 
subtraction  of  the  negative  number  ~oa? 

5.  How  does  the  result  of  subtracting  (5  a;  — 2  a;)  from  12  a; 
compare  with  the  result  of  subtracting  5  x  from  12  a;  ?  What 
effect  upon  the  result  has  the  subtraction  of  the  negative  number 
-2  a;? 

6.  Instead  of  subtracting  a  negative  number,  what  may  be 
done  to  obtain  the  same  result  ? 

62.  In  addition  two  numbers  are  given,  and  their  algebraic 
sum  is  required;  in  subtraction  the  algebraic  sum,  called  the 
minuend,  and  one  of  the  numbers,  called  the  subtrahend,  are 
given,  and  the  other  number,  called  the  remainder,  or  difference, 
is  required. 

Subtraction  is,  therefore,  the  inverse  of  addition. 
The  Difference  is  the  algebraic  number  that  added  to  the  sub- 
trahend gives  the  minuend. 

63.  Principles.  —  1.  Subtracting  a  positive  number  is  equiva- 
lent to  adding  a  numerically  equal  negative  number. 

2.  Subtracting  a  negative  number  is  equivalent  to  adding  a  numer- 
ically equal  positive  number. 


SUBTRACTION  33 

The  difference  of  similar  terms,  only,  can  be  expressed  in  one 
term. 

Prindple  1  may  be  established  as  follows : 

Let  m  represent  any  minuend  and  +s  any  positive  subtrahend. 
It  is  to  be  proved  that  m  —+s  =  m  +-s. 

§  62,  to  find  the  remainder  when  +s  is  subtracted  from  m  is  to  find  the 
algebraic  number  that  added  to  +s  will  give  m. 

Since  the  algebraic  sum  of  +s  and  -s  is  0,  by  the  Associative  Law  for 
Addition  the  algebraic  sum  of  +s  and  m  +-s  is  m  +  0,  or  m. 

Hence,  the  algebraic  number  that  added  to  +s  gives  m  is  m  +-s. 

.•.  m  —+s  =  m  +~s. 

Principle  2  may  be  established  as  follows : 

Let  m  represent  any  minuend  and  -s  any  negative  subtrahend. 

It  is  to  be  proved  that  m  —-s  =  »i  ++s. 

§  62,  to  find  the  remainder  when  -.s  is  subtracted  from  m  is  to  find  the 
algebraic  number  that  added  to  s  will  give  m. 

Since  the  algebraic  sum  of  -s  and  +s  is  0,  by  the  Associative  Law  for 
Addition  the  algebraic  sum  of  -s  and  m  ++s  is  m  +  0,  or  m. 

Hence,  the  algebraic  number  that  added  to  -s  gives  m  is  m  +  +s. 

.'.  m  —-S  —  m  ++S. 

64.  Since,  from  the  above  principles,  subtracting  algebraic 
numbers  is  equivalent  to  adding  them  to  the  minuend  with  their 
signs  changed,  it  follows  that  the  Laws  of  Order  and  Grouping 
for  Addition  hold  in  the  subtraction  of  algebraic  numbers;  and 
that  when  one  or  more  subtrahends  with  their  signs  changed  are 
added  to  the  minuend  to  form  the  algebraic  sum  called  the  dif- 
ference, one  set  of  signs,  +  and  — ,  suffices  to  denote  either  quality 
or  operation. 

65.  To  subtract  when  the  terms  are  positive. 

Examples 

1.     From  10  a;  subtract  4  a;. 

PROCESS 

y^  ^  Explanation.  —  Since  subtracting  a  positive  term  is  equiv- 

.  alent  to  adding  a  numerically  equal  negative  term  (IMn.  1), 

4  X  may  be  subtracted  from  10  x  by  changing  the  sign  of  4  x, 
and  adding  10  x  and  —  4  %. 


6x 


ALG.  — 3 


84 


SUBTRACTION 


2.    From  10  x  subtract  15  a;. 


PROCESS 

10  a; 
15  a; 


Explanation.  —  Since  subtracting  a  positive  term  is 
equivalent  to  adding  a  numerically  equal  negative  term 
(Prin.  1),  15x  may  be  subtracted  from  10  x  by  changing  the 
sign  of  15  X  and  adding  10  a;  and  —  15  x. 


3.  4.  5.  6.  7.  8. 

From        12  a        9  am        8ar/      24  mn^        CVo^      11  (a  +  6) 
Take  5  a      21am      18  .tV      12  mn^      15Vaa;      21  (a  +  &) 


From  9  a  +  7  6 

Take  2  a  +  3  6 


10. 

5a +  106 

7a  4-    Ah 


11. 

10x  +  2y 
Q>x  +  Ay 


12. 

3m  +  3n 
2  m  +  5  n 


13.  14.  15.  16. 

From       15m+n  1  x-\-2y  Ax  +  Ay  ^p  +  Zq 

Take       12m  +  2n  Ax-\-Ay  lx^2y         10p  +  2g 


17.  From  8jp  +  32  subtract  lOp  +  2. 

18.  From  15  m  +  rt  subtract  5  m  +  3  n. 

19.  From  'dax-\-^hy  subtract  Aax-\-^hy. 

20.  From  8  ahc  +  19  mx  subtract  20  ahc  +  7  rax. 

21.  From  a  +  3&  +  c  subtract  a  +  6  +  3c. 

22.  From  12  a^  +  2  6^  + 14  c^  subtract  3  a-  + 13  5^  -|-  3  &. 

23.  From  &  ax-\-hy  -\-l  cz  subtract  2ax-\-'by -\-2  cz. 

24.  From  7  ax-\-hy  +  2  cz  subtract  Aax-\-ohy  +  cz. 

25.  From  Aah  +  c  subtract  or -{■  Ir^ -\- ahc  +  2  ah -{■  2  c. 

26 .  From  5  xy  subtract  o;^  +  2  a;^?/  +  2  a;/  +  3  xi/  +  y . 

27.  From  1  subtract  a;*  +  13  ar*  + 15  a;^  +  16  a;  +  25. 

28.  From  Ix^  —  Ay^  subtract  6  a;^  +  3  a;^/ —  6  ?/l 


SUBTRACTION  35 

66.   To  subtract  when  some  terms  are  negative. 

Examples 
1.    From  8cc  — 3?/  subtract  5x  —  7y. 

PROCESS  Explanation.  —  Since  subtracting  a  positive  term  is  equiv- 

8  a; 3  y        alent  to  adding  a  numerically  equal  negative  term,  subtract- 

K     J  ing  5x  from  8  x  is  equivalent  to  adding  —  5x  to  8.r  (Prin.  1). 

Since  subtracting  a  negative  term  is  equivalent  to  adding 
a  numerically  equal  positive  term,  subtracting  —  7  y  from 


Sx-j-Ay        —3?/   is  equivalent  to  adding    +7y  to   —3y   (Prin.  2). 

Rule.  —  Change  the  sign  of  each  teryti  of  the  subtrahend,  or  con- 
ceive it  to  be  changed,  and  add  the  result  to  the  minuend. 

2.  3. 

From  5  a  6xy 

Take       -2a        -3xy 

7. 
From       Am  —  3n  +  2p 
Take       2m  —  5n  —    j) 

10. 
From         a  —  b-\-  c 
Take       2a  +  b  —  c 


4.                    5. 

—  Omn        -ISVx 

—  4  m7i        —  -SVa; 

6. 

-    3(a  +  &) 
-10(a  +  6) 

8. 

Sa-lOb  +  c 
6a—    5b  —  c 

9. 

3x  +  2y-z 
5x  —  Ay  —  z 

11. 

8a%-5ac^  +  9a^c 
3a^b-\-2ac'  +  9a^c 

12. 

r  —  s  +  t 
r  +  s  —  t 

13.  From  5x  —  3y-\-z  take  2x  —  y  +  %z. 

14.  From  3  a%  +  Z^'  -  a^  take  Aa%-8  a''  +  2  /A 

15.  From  rSa^  +  5b''-Ac^  take  Scr  +  Oft^  +  lOc^ 

16.  From  15x  —  3y  +  2z  subtract  3x  +  8y  —  dz. 

17.  From  a^  —  ab—b^  subtract  cib  —  2a'^  —  2b^. 

18.  From  m^  —  mn-\-n^  subtract  2  m^  —  3  mn -\- 2  n^. 

19.  From  5x^  —  2xy  —  y^  subtract  2  x^ -\-2xy  —  3y'\ 

20.  From  2ax  —  by  —  5xy  subtract  2by  —  2ax  —  3  xy. 


36  SUBTRACTION 

21.  From  2  a  +  c  subtract  a  —  h+c. 

22.  From  2m  +  n  subtract  n  —  2p. 

23.  From  x-\-y  subtract  3a— 4  + y, 

24.  From  2  a^  +  2  xy  subtract  x^  —  xy  —  y\ 

25.  From  2a  — 2d  subtract  a  —  h  +  c  —  d. 

26.  From  2  &  subtract  h  —  a  —  c  —  d. 

'iil.    From  a'^  +  .x'^  subtract  a^  —  3  a^ic  +  3  ax- —  a^. 

28.  From  a*  -\-\  subtract  1  —  a  +  a^  —  a''  +  a*. 

29.  From  the  sum  of  3a^  — 2a&  — 6^  and  3a6  — 2a^  subtract 
a^-ah-  b-. 

30.  From    3x  —  y  +  z   subtract   the   sum   of    x  —  Ay  +  z    and 
2x  +  3y-2z. 

31.  From   a  +  b  +  c  subtract  the  sum  of  a  —  b  —  c,  b  —  c  —  a, 
and  c  —  a  —  b. 

32.  Subtract  the  sum  of  mhi  —  2  mn^  and  2  m^n  — m^— n^4-2  mn^ 
from  irv^  —  n^. 

33.  Subtract  the  sum  of  2c  —  9a  —  36  and  36  —  5a  —  5c  from 
h  —  3c  +  a. 

34.  From    3 6.1;  +  4 oy   subtract   the   sum   of    Say  —  Abx    and 
bx  -\-  ay. 

35.  From  the  sum  of  1  -\-x  and  1  —  a^  subtract  1  —  a;  +  a;-  —  x*^. 

36.  From  |a;»— far'  +  3x  — 7  subtract  ^x^  —  ^xr  +  ^x  —  lO. 

37.  From  i  m^  —  \  mhi  + 1  ??irr  —  27  ^^  subtract  n^  —  m^  +  ^  mn^ 

38.  From    5{a  +  b) —3(x-\-y) +4:(m  +  n)    subtract    4 (a +6) 

+  2(x  +  y)  +  {m  +  n). 

39.  From    n^  —  m^    subtract  the   sum   of    2  mhi^  —  3  mn*  and 
m^  +  4  m^n^  —  2  m^n^  +  5  mrt*  —  n^. 

40.  From  the  sum  of  3a^  —  2a;  +  l   and  2x  —  5  subtract  the 

sum  of  a;  —  a:^  + 1  and  2  a^  —  4  x  +  3. 


SUBTRACTION  37 

PARENTHESES 

67.  The  subtrahend  is  sometimes  written  within  a  sign  of 
aggregation  preceded  by  the  sign  — . 

If  a  —  6  is  to  be  subtracted  from  2  a,  it  may  be  written  2  a  —  (a  —  ft). 

1.  What  change  must  be  made  in  the  signs  of  the  terms  of  the 
subtrahend,  when  it  is  subtracted  from  the  minuend  ? 

2.  When  a  number  in  parenthesis  is  preceded  by  the  sign  — , 
what  change  must  be  made  in  the  signs  of  the  terms,  when  the 
subtraction  is  performed,  or  when  the  parenthesis  is  removed  ? 

68.  Principles.  —  1.  A  parenthesis  preceded  by  the  minus  sign 
may  he  removed  from  an  expression,  if  the  signs  of  all  the  terms  in 
pareyithesis  are  changed. 

2.  A  parenthesis  jtreceded  by  the  minus  sign  may  be  used  to  inclose 
an  expression,  if  the  signs  of  all  the  terms  to  be  inclosed  are  changed. 

1.  When  numbers  are  inclosed  in  a  parenthesis  preceded  by  the  plus 
sign,  the  parenthesis  may  be  removed  without  changing  the  signs  of  the 
terms. 

2.  Any  number  of  terms  may  be  inclosed  in  a  parenthesis  preceded  by  » 
plus  sign  without  changing  the  signs  of  the  terms. 

3.  The  student  should  remember  that  in  an  expression  like  —  (x  —  ?/),  or 
—  a;  —  y,  the  sign  of  a;  is  plus,  and  the  expression  is  the  same  as  if  it  were 
written  —  (+  x  —  y),  or  — v  x  —  y. 

Examples 
Simplify  the  following : 

1.  tt  +  (6  — c).  8.  a  — 6  — (c  — d). 

2.  a  —  ip—c).  9.  a  — &  — (— c  +  a). 

3.  X  —  (y  —  z).  10.  a  —  m  —  (n  —  m). 

4.  x-{-y-\-z).  11.  5a- 2&- (a -26). 

5.  m  —  n  —  (—a).  12.  a  —  (6  —  c  +  a)  — (c  —  6). 

6.  m  —  (n  —  2a).  13.  2xy -\- 3y^  —  (x^ +  xy —  y^. 

7.  5x  —  {2x  +  y).  14.  m  +  (3  m  —  n)  —  (2 ri  —  m)  +  n. 


38  SUBTRACTION 

Collect  in  alphabetical  order  the  coefficients  of  x  and  y  in  the 
following,  giving  each  parenthesis  the  sign  of  the  first  coefficient 
to  be  inclosed  therein : 

15.    ax  —  by  —  3bx-\-2cy—fx  —  gy. 

PROCESS 

cue  —  by  —  3  bx  +  2  cy  —  fx  —  gy  =(a  —  3  b—f)x—(b  ~2c-\-  g)y 

Suggestion. — The  coefficient  of  y  is  —  b  +  2c  —  g,  which  is  written 
-(6  -2c  +  gr)  (Prin.  2). 

16.  ax  —  by  —  bx  —  cy  -{- dx  —  ey.  21 .  bx  —  cy  —  2ay  ■+-  by. 

17.  mx  —  2  ny  +  nx  —  ry  —  px  +  qy.  22.  mx  —  bx  —  4y  —  my. 

18.  5  ax  +  3  ay  --  2  dx  -{-  ny  —  5  X  —  y.  23.  rx  —  ay  —  sx-\-2  cy. 

19.  cx~2bx  +  7  ay -\-3ax —  Ix —  ty.  24.  x^ -\- ax  —  y^  +  ay. 

20.  bx+  cy  —  2 ax  +  by  —  ex  —  dy.  25.  a?  —  ay  —  ax  —  y^. 

Group  the  same  powers  of  x  in  the  following : 

26.  an?  +  ba?  —  cx-\-  ea?  —  dv?  —  fx. 

27.  x'-^3:)?^3x-ax'-3a'3?^-bx. 

28.  a^  —  abx  —  x?  —  bx^  —  ex  —  mnx^  +  dx. 

29.  ax'^  —  X*  —  aoi?  -\-  x^  -\- ax  —  x  —  abx?  +  y?. 


30.    Simplify  2  a-\a-\b  -  i^b  -2  a  -  b)\'\-(b  -  a). 

When  an  expression  contains  parentheses  within  parentheses, 
they  may  be  removed  in  succession,  beginning  with  either  the 
outermost  or  the  innermost,  preferably  the  innermost. 


Solution 


2a-[a-{6-(3  6-2a-  6)}]  -(b-a) 

Prin.  1,  =2a~la-{b-(3b-2a  +  b)}]-b  +  a 

Uniting  terms,  =3d—  [a  —  {6—  (4  6  —  2  a)}]  —  b 

Prin.  1,  =  3  a  -  [a  -  {6  -  4  6  +  2  a}]  -  6 

Uniting  terms,  =3a—  [a  —  {— 3&  +  2a}]— 6 

Prm.  1,  =3a-[a  +  36-2a]-6 

Uniting  terms,  =3a— [— a  +  3  6]— 6 

Prin.  1,  =Sa  +  a-Sb  —  b 

Uniting  terms,  =  4  a  —  4  6. 


SUBTRACTION  39 

Simplify  the  following : 

31.  ^a  +  h  —  \x  +  4.a  +  h  —2y  —  (x  +  y)\. 

32.  ah  —  \ab  -\-  ac  —  a  —  (2 a  —  ac)  -{■  {2 a  —  2 ac)\. 

33.  a-^[t/_|5+4a-(62/  +  3)^-(72/-4a-l)]. 

34.  4m— [2) +  3?i  — (m  +  n) +3  —  (6p  — 3n  — 5m)]. 


35.  a  +  26  +  (14a-56)-^6a  +  6&-(5a-4a-46)|. 

36.  12a-{[4-36-(6&  +  3c)]+6-8-(5a-2&-6)(. 

37.  a+6— {— [a+&  — (c+i«)]  — [3a— (c— a;+a)-&]+4a|. 

38.  x^-lx'-il-x)]-]!  +[_:>^ -  (1  - x)  +  ar'J |. 


39.  4  -  [ [5  y  -  (3  -  2  a;  -  2)]  -  [a:  +  (5  y  -  a;  +  3)]  I . 

40.  ah  -\b  +  X  -  {h  +  c  -  ah  +  x)\+[_x  -  {h  -  c  -  7)]. 

41.  a-  -  6^  _  ^ad  +  a^  -  (a;  +  a-  -  h^)  -W\  +  5  ad  -  (x  +  3  ad). 

42.  a-(Z>-c)-[a-{6— c-(6+c-a)  +  (a-&)  +  (c-a)|]. 

43.  —  f3ax-[5a;r/-32;]  +  2  —  (4a;// +[6z  +  7aa;]-f  3  2;)*. 

44.  1  -  X  -\1  —  X  -\\  —  X  —  {1  —  x)  —  {x  —  1)']—  X  -{-\\. 

45.  1  -a;  — {1  -[a;-l  +  (a;-l)  —  (1  — »)  — a;]+l-a;|. 

46.  a;-[— {- (-a;) +  a;|  —  2a;]. 

47.  (a  —  6)  -  {  -  a  -  (6  —  a)  +  (a  —  &)}. 


48.  a  — 7— [— {  —  a  — (—a —  a —  3)1]. 

49.  a  —  «  — [— {a  +  (a;  —  a)  —  (a;  —  4a)|]. 


50.    bxy-l-\{y''-xy)~(xy-y''-2xy)\'\. 


51.    2a-[a- {&-(36~2a-6)n-(6-a)- 


52.  a  — [— (m  — a)  —  Ja  —  (m  —  2m  +  6a)|]. 

53.  a  —  \-h-  (c-d)^  +  a-[-&  +  ^-2c-  (d-e)f]. 


54.    a2  +  5-[2a&-[-(7-3a&)  -ab  +  2d' -  z\- {Za- z)^ 


55.    2  a;  +  (3  ?/  —  52  X  -  [?/  +  4  a;  -  (3  y  -  a;)]  —  2  2/1  -  a;  —  ?/). 


56.    l-(-{-[-(-a-a-l)-3]-2|-a)-[a-(a-l)]. 


40  SUBTRACTION 

TRANSPOSITION  IN  EQUATIONS 

69.    1.  What  number  diminished  by  2  is  equal  to  8  ? 

2.  If  a  number  increased  by  2  is  equal  to  8,  what  is  the 
number  ? 

3.  In  the  equation  a;  —  2  =  8,  what  is  done  with  the  2  in 
obtaining  the  value  of  x  ?  In  the  equation  ic  =  8  +  2,  how  does 
the  sign  of  2  compare  with  its  sign  in  the  previous  equation  ? 

4.  In  the  equation  a;  -}-  2  =  8,  what  is  done  with  the  2  in 
obtaining  the  value  of  x  ?  In  the  equation  cc  =  8  —  2,  how  does 
the  sign  of  2  compare  with  its  sign  in  the  previous  equation  ? 

5.  In  changing  the  2's  from  one  side  of  the  equation  to  the 
other,  what  change  was  made  in  the  sign  ? 

6.  When  a  term  is  changed  from  one  side,  or  member  of  an 
equation,  to  the  other,  what  change  must  be  made  in  its  sign  ? 

7.  If  3  is  added  to  one  member  of  the  equation  2  +  5  =  7, 
what  must  be  done  to  preserve  the  equality? 

8.  If  3  is  subtracted  from  one  member  of  the  equation 
2  +  8  =  10,  what  must  be  done  to  preserve  the  equality  ? 

9.  If  one  member  of  the  equation  2  +  5  =  7  is  multiplied  by 
5,  what  must  be  done  to  preserve  the  equality  ? 

10.  If  one  member  of  the  equation  10  +  25  =  35  is  divided  by 
5,  what  must  be  done  to  preserve  the  equality  ? 

11.  If  one  member  of  the  equation  a;  =  5  is  raised  to  the  second 
power,  what  must  be  done  to  preserve  the  equality  ? 

12.  If  the  square  root  of  one  member  of  the  equation  x'  =  25 
is  taken,  what  must  be  done  to  preserve  the  equality? 

13.  What,  then,  may  be  done  to  the  members  of  an  equation 
without  destroying  the  equality  ? 

70.  The  parts  of  an  equation  on  each  side  of  the  sign  of 
equality  are  called  its  Members. 

The  part  on  the  left  of  the  sign  of  equality  is  called  the  First 
Member,  and  the  part  on  the  right,  the  Second  Member. 


SUBTRACTION  41 

71.  The  process  of  changing  a  term  from  one  member  of  an 
equation  to  the  other  is  called.  Transposition. 

72.  Principle.  —  A  term  may  he  transposed  from  one  member 
of  an  equation  to  the  other,  provided  its  sign  is  changed. 

73.  A  truth  that  does  not  need   demonstration  is  called  an 
Axiom. 

74.  Axioms.  —  1.    Things  that  are  equal  to  the  same  thing  are 
equal  to  each  other. 

2.  If  equals  are  added  to  equals,  the  sums  are  equal. 

3.  If  equals  are  subtracted  from  equals,  the  remainders  are  equal. 

4.  If  equals  are  multiplied  by  equals,  the  products  are  equal. 

5.  If  equals  are  divided  by  equals,  the  quotients  are  equal. 

6.  The  same  powers  of  equal  numbers  are  equal. 

7.  Tlie  same  roots  of  equal  numbers  are  equal. 

Equations  and  Problems 

75.  1 .    li  bx  —  2  =  'dx  +  &,  find  the  value  of  x. 

PROCESS  Explanation.  —  Since  the  value  of  x  is  sought, 

5  ^.  _  2  =  3  aj  -J-  6      ^'^  terms  containing  x  must  be  collected  in  one 


Sx         =  3  X 


member  of  the  equation,  and  the  remaining,  or 
known  terms  in  the  other  member. 
^  ^      ^  ^  '"  3x  may  be  made  to  disappear  from  the  second 

2  =  2 member  by  subtracting  3x  from  both  members. 

2  X  =8  The  result  (Axiom  .3)  is  the  equation  2  x  —  2  —  Q. 

.    /J.  __  4.  —  2  may  be  made  to  disappear  from  the  first 

member  by  adding  2  to  both  members.     The  result 
(Axiom  2)  is  the  equation  2  a;  =  8. 
o  X  —  A     =oa;  +  D  Dividing  both  members  of  this  equation  by  2, 

5a;  —  3a:  =  24-6         the  coefiicient  of  x,  the  resulting  equation  (Axiom 
2  x  =  8  5)  is  X  =  4,  the  value  of  x  sought. 

_ .  x  =:  4  ^^'  since  a  term  may  be  transposed  from  one 

member  to  the  other  if  its  sign  is  changed  (Piin.), 
VERIFICATION  3  x  transposed  to  the  first  member  becomes  —  3  x 

20 2  ^  12  +  6       ^°^  ~  ^  transposed  to  the  second  member  becomes 

+  2.     Therefore,  the  resulting  equation  is 
18  ^  lo 

5x-3x  =  2  +  6. 

Uniting  terms,  2  x  =  8.     Dividing  both  members  by  2,  the  result  is  x  =  4. 


42  SUBTRACTION 

The  result  is  verified  by  substituting  the  value  of  x  for  x  in  the  original 
equation.  If  the  members  are  then  identical,  tlie  value  found  for  the  un- 
known number  is  correct. 

2.   If  5  a;  -  7  =  30  -  7,  find  the  value  of  x. 

PROCESS 

5  a;  -  7  =  30  -  7. 
5x  =  30. 
.-,  ic  =  6. 

Suggestion.  — Since,  if  —  7  were  transposed  from  the  first  member  to  the 
second,  it  would  appear  as  +  7  and  cancel  the  term  —  7  in  that  member,  the 
two  equal  terms  may  be  canceled  before  the  transposition. 

RuLK.  —  Transpose  terms  so  that  the  unknoivn  terms  stand  in  the 
first  member  of  the  equation  and  the  knovm  terms  in  the  second. 

Unite  similar  terms,  and  divide  both  members  of  the  equation  by 
the  coefficient  of  the  unknown  number. 

Verificatiox.  —  Substitute  in  the  original  equation  the  value  of 
the  unknoion  number  thus  found.  If  the  members  of  the  equation 
are  then  identical,  the  value  of  the  unknown  number  found  is  correct. 

1.  The  same  term  with  the  same  sign  in  both  members  of  an  equation 
may  be  canceled  (Ax.  2  or  3). 

2.  If  the  signs  of  all  the  terms  of  an  equation  are  changed,  the  equality 
will  not  be  destroyed;  for  (Ax.  3)  both  members  may  be  subtracted  from 
0  without  destroying  the  equality. 

Find  the  value  of  x  and  verify : 

3.  5  07  + 3  =  8.  10.  7  0^-10  =  60. 

4.  iK  +  5  =  ll.  11.  7 +  2  a;  =  11. 

5.  a;- 5  =  11.  12.  3  +  2 a;  =  15. 

6.  2  37-3  =  21.  13.  1  +  12 a;  =  85. 

7.  oa;  +  7  =  42.  14.  5  +  3.^  =  11. 

8.  3a;- 2  =  25.  15.  7  +  5.x'  =  47. 

9.  2a;  +  4  =  10.  16.  2 -|- 9  x  =  74, 


SUBTRACTION  43 

17.  3  =  5-a;.  26.  7  a; -12  =  a; +  13 -|- 5. 

18.  9-5x  =  -l.  27.  4a;-20  =  5iC-50  +  a;. 

19.  7a;  +  2  =  a;  +  14.  28.  3  a;  +  16  =  20 -5  a; +  4. 

20.  oa;  — 5  =  2x  +  4.  29.  7  a;  -  55  =  18  —  2a;  — 1. 

21.  3a; +  2  =  a; +  30.  30.  -a; -12  =  40 -8 a; +  4. 

22.  5a;- 2  =  2a; +  7.  31.  80 -3 a;  =  83 -8 a; +  7. 

23.  2 +  13  a;  =  50 -9.  32.  9  a; -90  =  16 -a; +  4. 

24.  10  + a;  =  18 -a;.  33.  50 -a;  =20  + a;. 

25.  2a; +  2  =  32 -a;.  34.  7  a; +  25  =  30 +  6 a; -3. 

Solve  the  following  problems  : 

35.  What  number  increased  by  10  is  equal  to  19  ? 

36.  What  number  diminished  by  30  is  equal  to  20  ? 

37.  What  number  diminished  by  111  is  equal  to  —  15? 

38.  What  number  exceeds  \  of  itself  by  10? 
Suggestion.  — Let  3  as  =  the  number. 

39.  Five  times  a  number  exceeds  3  times  the  number  by  14. 
What  is  the  number  ? 

40.  If  5  is  subtracted  from  a  certain  number,  and  the  differ- 
ence is  subtracted  from  3  times  the  number,  the  result  is  35. 
What  is  the  number  ? 

41.  The  double  of  a  number  is  64  less  than  10  times  the  num- 
ber.    What  is  the  number  ? 

42.  If  4  is  subtracted  from  a  certain  number,  and  the  differ- 
ence is  subtracted  from  40,  the  result  is  3  times  the  number. 
What  is  the  number  ? 

43.  Three  times  a  certain  number  is  as  much  less  than  72  as 
4  times  the  number  exceeds  12.     What  is  the  number  ? 

44.  Twice  a  certain  number  exceeds  \  of  the  number  as  much 
as  6  times  the  number  exceeds  65.     What  is  the  number  ? 

45.  If  16  is  added  to  a  certain  number,  the  result  is  5Q  dimin- 
ished by  7  times  the  number.     What  is  the  number  ? 


44  SUBTRACTION 

46.  If  6  times  a  certain  number  lacks  as  much  of  62  as  3 
times  the  number  exceeds  19,  what  is  the  number  ? 

47.  Three  times  a  certain  number  increased  by  a  is  equal  to 
the  number  increased  by  9  a.     What  is  the  number  ? 

48.  The  sum  of  4  numbers  in  a  row  is  58.  If  their  common 
difference  is  3,  what  are  the  numbers  ? 

Suggestion.  —  Let  x  =  the  smallest  number. 

Then,  at  +  3  =  the  second  number, 

X  +  6  =  the  third  number, 
and  X  +  9  =  the  fourth  number. 

49.  A  man  distributed  1  dollar  among  5  boys  so  that  each  boy 
except  the  youngest  received  5  cents  more  than  the  boy  next 
younger.  If  the  boys  were  all  of  different  ages,  how  much  did 
each  receive  ? 

50.  The  common  difference  of  5  numbers  is  2,  and  their  sum 
is  100.     What  are  the  numbers  ? 

51.  John  and  James  were  comparing  their  earnings.  John 
said,  "  I  have  earned  50  cents."  James  replied,  "  If  I  had  earned 
half  as  much  as  I  have,  and  10  cents  more,  I  should  have  earned 
the  same  as  you."     How  much  had  James  earned? 

52.  A  drover,  when  asked  how  many  cattle  he  had,  replied,  "  If 
I  had  \  more  than  I  have  and  2  more,  I  should  have  200."  How 
many  cattle  had  he  ? 

53.  The  earnings  of  a  mill  for  4  years  were  $46,000.  If  the 
books  showed  an  annual  increase  of  $  1000,  what  were  the  earn- 
ings for  each  year  ? 

54.  James  had  |  as  much  money  as  John,  John  5  cents  less 
than  William,  and  Robert  5  cents  more  than  3  times  as  much  as 
James.     If  they  together  had  $  1.50,  how  much  had  each  ? 

Suggestion.  — Let  3  x  =  the  number  of  cents  John  had. 

55.  A  speculator  who  doubled  his  money  by  a  fortunate  invest- 
ment, afterward  lost  f  600,  but  he  still  had  $  400  more  than  the 
original  sum.     How  much  had  he  at  first  ? 


MULTIPLICATION 


76.   1.    How  many  are  3x  +  3x  +  3a;  +  3a;  +  3a;? 

2.  How  many  are  5  times  3x?     2  times  3a;?     7  times  2a? 

3.  A  man  saves  f  10  a  month,  indicated  by  +10.  How  many 
dollars  will  he  save  in  a  year  ?  What  sign  should  be  placed  be- 
fore the  result  to  indicate  the  number  of  dollars  saved  ? 

4.  How  many  are  12  times  +10?  12  times  +10  a?  5  times 
+3x?     8  times  +2 m ? 

5.  When  a  positive  number  is  multiplied  by  a  positive  num 
ber,  what  is  the  sign  of  the  product  ? 

.6.  If  a  man  loses  $5  a  month,  indicated  by  "5,  how  many 
dollars  will  he  lose  in  a  year  ?  What  sign  should  be  placed  be- 
fore the  result  to  indicate  the  number  of  dollars  lost  ? 

7.  How  many  are  12  times  "5?  12  times  "5  6?  7  a  times 
-3x?     3  times  '2  b  ?     11  times  Sy?     5  times  "5  ab  ? 

8.  When  a  negative  number  is  multiplied  by  a  positive  num- 
ber, what  is  the  sign  of  the  product  ? 

9.  If  a  man's  gains  in  business  are  $10  a  month,  indicated 
by  +10,  how  many  dollars  less  had  he  3  months  ago,  indicated  by 
~3,  than  he  has  now  ?     Indicate  the  result  algebraically. 

10.  How  many  are  +10  multiplied  by  ~3  ?  +10  multiplied  by 
-f)  ?     +2  multiplied  by    3  ?     +a  multiplied  by  -&  ? 

11.  What  is  the  sign  of  the  product  when  a  positive  number 
is  multiplied  by  a  negative  number  ?  When  a  negative  number 
is  multiplied  by  a  positive  number  ? 

12.  What,  then,  is  the  sign  of  the  product  of  two  numbers 
having  imlike  signs  ? 

45 


46  MUL  TIPL ICA  TION 

13.  If  a  man  who  is  in  debt  is  getting  deeper  in  debt  at  the 
rate  of  $10  a  month,  indicated  by  ~10,  how  much  better  off  was 
he  3  months  ago,  indicated  by  ~3,  than  he  is  now  ?  Indicate  the 
result  algebraically. 

14.  How  many  are  ~10  multiplied  by  ~3?  ~10  multiplied  by 
-5  ?     -2  multiplied  by  "3  ?     "a  multiplied  by  'h  ? 

15.  What  is  the  sign  of  the  product  of  two  negative  numbers  ? 
of  two  positive  numbers  ? 

16.  What,  then,  is  the  sign  of  the  product  of  two  numbers 
having  like  signs  ? 

17.  What  is  the  sign  of  "5  x  "2  ?  of  "5  x  "2  x  "2  ?  of  "5  x  "2 
X+2X-2?  of  +5x-2x-2x  2x-2?  of  "2  x-3  x+2  x+2  x-2 
X-2X-2? 

18.  What  sign  has  a  product,  if  the  number  of  negative  fac- 
tors is  evenf  What  sign  has  a  product,  if  the  number  of  negative 
factors  is  odd? 

19.  What,  then,  determines  the  sign  of  a  product? 

20.  In  the  expression  a?,  what  is  3  called  ?  What  does  it  indi- 
cate ?     In  a^  how  many  times  is  a  used  as  a  factor  ? 

21.  When  a?  is  multiplied  by  a*,  how  many  times  is  a  used  as 
a  factor  in  the  product  ?  when  a^  is  multiplied  by  a^  ? 

When  a"  is  multiplied  by  a",  what  is  the  product,  if  m  and  n 
are  positive  integers  ? 

22.  How,  then,  is  the  exponent  of  a  factor  in  the  product  de- 
termined ? 

77.  When  the  multiplier  is  a  positive  integer,  the  process  of 
taking  the  multiplicand  additively  as  many  times  as  there  are 
units  in  the  multiplier  is  called  Multiplication. 

When  the  multiplier  is  any  number,  multiplication  may  be 
defined  as  the  process  of  finding  a  number  that  has  the  same  rela- 
tion to  the  multiplicand  as  the  multiplier  has  to  1. 

The  multiplicand  and  multiplier  are  called  the  factors  of  the 
product. 

78.  Principles.  —  1.  Law  of  Signs.  —  The  sign  of  the  prod^ict 
of  tico  factors  is  +  when  they  have  like  signs,  and  —  when  they 
have  unlike  signs. 


MUL  TI PLICA  TION  47 

2.  Law  of  Coefficients.  —  The  coefficient  of  the  product  is  equal  to 
the  2>roduct  of  the  coefficients  of  the  factors. 

3.  Law  of  Exponents.  —  The  exponent  of  a  number  in  the  prod- 
uct is  equal  to  the  sum  of  its  exponents  in  the  factors. 

79.   The  Law  of  Signs  may  be  established  as  follows : 

+3=+l++l++l,  (1) 

and  -3=-l4.-l +-1  =_+l  _+l  _+l;  (2) 

that  is,  +3  is  obtained  from  +1  by  taking  +1  additively  three  times,  and    3 
by  taking  +1  subtractively  three  times. 

Hence,  §  77,  multiplying  any  number  by  +3  is  equivalent  to  taking  that 
number  additively  three  times,  and  multiplying  any  number  by  -3  is  equiva- 
lent to  taking  that  number  subtractively  three  times. 

By  (I),  +5  X +3  =+5 ++5 ++5  =+15,  (3) 

and  5  x+3=-5+-5+-5=-15.  (4) 

By  (2),  +5  x-3  =  -+5-+5-+5=-15,  (5) 

and  -5  X -3  =--5 --5 --5  =+15.  (6) 

Similarly,  since +(|)  =  +  (i)  +  +  a)++Q)  and -(|)=-+(|)-+(|)-  +  Q), 
+5x+(|)  =  +  (f)  ++(f)  +  +  (|)=  +  (^);  (7) 

and  so  on,  as  in  (4),  (5),  and  (6). 

In  (3)  and  (6)  the  product  of  two  algebraic  numbers  with  like  signs  is 
positive. 

In  (4)  and  (5)  the  product  of  two  algebraic  numbers  with  unlike  signs  is 
negative. 

(7)  shows  that  like  results  are  obtained  when  the  multiplier  is  a  fractional 
number  (§  7). 

Passing  to  general  symbols,  let  a  and  h  be  any  absolute  numbers. 
First,  when  h  is  a  whole  number. 


Since 

+6=+l  ++1  4-+1  +  ••• 

to  b  terms,. 

§77, 

+a  x+b  =+a  ++a  ++a  +  •• 

•  to  b  terms 

=+ab, 

(8) 

and 

-a  x+b  =-a  +-a  +-a  +  ••■ 

•  to  b  terms 

=   a6. 

(9) 

Since 

-b  =  -+l  -+1  -+1  - 

•  ••  to  b  terms, 

§77, 

+a  x-b  =  -+rt  — +a  — +a  — 

'••  to  b  terms 

=-ab, 

(10) 

and 

-a  x-b  =  —-a  —~a  —~a  — 

•  ••  U)b  terms 

=  +ab. 

(11) 

48  MUL  TIPLICA  TION 

Second,  when  b  is  a  fractional  number. 

As  in  (7),  the  same  reasoning  applies  when  6  is  a  fractional  number. 

Hence,  from  (8)  and  (11),  the  product  of  any  two  algebraic  numbers  with 
like  signs  is  positive  ;  and  from  (9)  and  (10),  the  product  of  any  two  algebraic 
numbers  with  unlike  signs  is  negative. 

When  the  multiplier  is  a  positive  or  negative  whole  or  fractional 
number,  it  appears  from  the  above  proofs  that  algebraic  multi- 
plication is  only  abbreviated  algebraic  addition.  Hence,  as  in 
addition,  but  one  set  of  signs  +  and  —  is  required  to  denote  both 
quality  and  operation. 

Hence,  the  Law  of  Signs  may  be  expressed  as  follows : 

+  a  multiplied  by  +  6  =  +  ab, 

—  a  multiplied  by  —  b  =  +  ab, 

—  a  multiplied  hy  -\-  b  =  —  ab, 
and                         +  a  multiplied  by  —  b  —  —  ab. 

80.  It  follows  from  the  Law  of  Signs,  applied  repeatedly,  that 
the  ^noduct  of  any  number  of  algebraic  numbers  is  +  when  the 
number  of  negative  factors  is  even,  and  —  when  the  number  of 
negative  factors  is  odd. 

81.  The  Law  of  Exponents  or  the  Index  Law  for  multiplication 
may  be  established  for  positive  integral  exponents  as  follows : 

Let  m  and  n  be  any  positive  integers. 
By  the  definition  of  a  power,  §  24, 

a'^  —  a  X  a  X  a  ...  to  m  factors, 
a^  =  a  X  a  X  a  ...  to  n  factors  ; 
.*.  a™  X  a"  =  (a  X  a  X  rt  ...  to  Tw  factors)  (a  x  a  x  a ...  to  n  factors) 
=  «  X  a  X  a  ...  to  (m  +  n)  factors. 
Hence,  a™  xa"=  «"»+". 

In  like  manner,  a'^  x  a*^  x  a^  =  a'"+"+p. 

Thus,  a^x  a*  =  a^+*  =  a^ 

and  a3  X  a2  X  a*  =  a^+^+*  =  a^. 

82.  1.  How  does  2x5  compare  with  5  x  2  in  value?  3x7 
with  7x3?     2x5x6  with  2x6x5? 

2.  What  is  the  effect  upon  the  value  of  a  product  of  changing 
the  order  of  its  factors  ? 


MUL  TIPLICA  TION  49 

Law  of  Order,  or  Commutative  Law  for  Multiplication.  —  The  fac- 
tors of  a  product  may  be  taken  in  any  order. 

The  Law  of  Order  may  be  established  as  follows : 

Since  the  number  of  negative  factors  will  not  be  changed  by  taking  the 
factors  in  any  order,  §  80,  the  sign  of  the  product  is  the  same  in  whatever 
order  the  factors  are  taken. 

We  know  from  arithmetic  that  arithmetical  numbers  may  be  multiplied 
in  any  order.  Hence,  the  absolute  value  of  the  product  is  the  same  in  what- 
ever order  the  factors  are  taken. 

Since  neither  the  sign  nor  the  absolute  value  of  the  product  of  algebraic 
numbers  is  changed  by  changing  the  order  of  the  factors,  the  factors  may  be 
taken  in  any  order. 

In  general  symbols,  a  x  b  x  c  x  ••'  =  b  x  c  x  a  x  •••  =  etc. 

83.  1.  How  does  2  x  3  x  5,  or  6  x  5,  compare  in  value  with 
2  X  (3  X  5),  or  2  x  15  ?  with  5x6?  a  x  6  X  c,  or  (ab)  x  c, 
with  a  X  (be)  ? 

2.    How  may  the  factors  of  a  product  be  grouped  ? 

Law  of  Grouping,  or  Associative  Law  for  Multiplication.  —  The  fac- 
tors of  a  jyroduct  may  be  grouped  in  any  manner. 

The  Law  of  Grouping  may  be  established  as  follows  : 

By  the  notation  of  multiplication,  abc  denotes  that  a  is  to  be  multiplied 
by  b  and  then  the  product  ab  is  to  be  multiplied  by  c  ;  that  is, 

abc—(ab)c.  (1) 

1.  Let  it  be  required  to  prove  that  (ab)c  =  a(bc). 
By  the  Law  of  Order,  abc  =  bca 

by  notation,  =  (bc)a 

by  the  Law  of  Order,  =  a{bc).  (2) 

From  (1)  and  (2),  (ab)c  =  a(bc). 

Similarly,  it  may  be  proved  that      (ab)c  —  b(ac),  etc. 

2.  Let  it  be  required  to  prove  that  abed  =  (aft)  (cd). 
By  notation,  abed  =  ab  x  c  x  d 

putting  m  for  ab,  =  m  ■  c  •  d,  or  mcd. 

By  1,  m  •  c  •  d  =  m(cd). 

Putting  ab  for  m,  (ab)  •  c-  d  =  (ab)  (cd) ; 

that  is,  by  notation,  abed  =:(ab)(cd). 

Similarly,  it  can  be  shown  that  (abc)d  =  a(bed)  =  (he)  (ad)  =  (ac) (bd)  = 
(abd)c  =  (adc)b  =  c(dba)=  etc.,  the  factors  being  grouped  in  any  manner 
whatever, 

ALG.  — 4 


50  M  UL  T I  PLICA  TION 

3,  In  a  similar  way  tlie  law  may  be  established  for  any  number  of  factors, 
successively  for  5,  6,  7,  •••  factors. 

Hence,  ahc  •■■p  =  a{bc  •••p)=  b{ac  ■••p),  etc.,  for  all  values  of  the  letters. 
84.    To  multiply  a  monomial  by  a  monomial. 

Examples 
1.    Multiply  5  a^y^  by  —  3  xyh. 
„„^^„„„  Explanation.  —  Since  the  multiplier  is  composed  of  the 

J,      ,  „        factors  —  3,  x,  y'^,  and  z,  the  multiplicand  may  be  multiplied 
J^        by  each  successively.     —  3  times  5  x-y^  =  —  15  x^ij'^  (Prin.  1 
~        ^^        and  2);  x  times   -  15  xhf^  =  -  15  x^y^  (Prin.  3);  y^  times 
—  15  a^y^z        ~  IS  x^y^  =  -  15  xV  (Prin.  3)  ;  and  this  multiplied  by  z  is 
equal  to  —  I5x^y^z  (Prin.  3). 
Or,  since  the  signs  of  the  numbers  are  unlike,  the  sign  of  their  product 
is  —  ;  the  coeflBcient  of  the  product  is  the  product  of  the  coefficients  5  and  3  ; 
and  the  product  of  the  literal  numbers  is  expressed  by  affecting  each  with 
an  exponent  equal  to  the  sum  of  its  exponents  in  the  factors. 

Rule.  —  To  the  product  of  the  numerical  coefficients  annex  the 
letters,  each  with  an  exponent  equal  to  the  s%im  of  its  exponents  in 
both  factors. 

Write  the  sign  +  before  the  product  when  its  factors  have  like 
signs,  and  —  when  they  have  unlike  signs. 

2.  3.  4.  5.  6. 

Multiply     -2  6  -7  2  a  2m^ 

By  8  -2  -9  5  6m^ 


7. 

8. 

9. 

10. 

11. 

Multiply 

10  a" 

x-y'^           — 

4  abc 

5  a^b(^ 

-  2  xy^ 

By 

5  a' 

xif 

2a^b 

■  7  ab'c 

2x'y 

12. 

13. 

14. 

15. 

16. 

Multiply 

-Sa'x" 

—  5  im?n^ 

-  6  a^fiVa; 

4  abed 

-3x%f 

By 

-2  ax' 
17. 

3  mn 

—  4  a'bny^     ■ 
19. 

-1 

-1 

18. 

20. 

21. 

Multiply 

-  2  c^-^ 

-  3  n'y 

4  a^xb'y* 

-1       - 

5  mhi^dhf 

By 

—  4  ax* 

6b'y 
23. 

3  a^x^b^y 
24. 

-1       - 
25. 

2m'«u«c-y 

22. 

26. 

Multiply 

5  p(f\)? 

10  mV 

—  2  a^m'n* 

x-yz 

-phX-(i 

By 

-2rq*x 

-    Sn^'m* 

8  b'n^ni' 

—  x-yz^ 

—  ahc 

MULTIPLICATION 


61 


27. 

28. 

29. 

30. 

31. 

Multiply 

By 

2a'»+' 
32. 

—  5  a;" 

X 

-  x"y" 
3xy        - 

3,n-y-2 

xy 

4  a;"-' 

-2a;"+* 

33. 

34. 

35. 

36. 

Multiply 
By 

by 
-3.v"-2 

—  a" 
-a" 

a" 
38. 

-a;'- 
—  a;" 

ym—n+l 

37. 

39. 

Multiply 

a-b^x^y""^ 

a"-i6"-V 

m''n'b^y'' 

By 

QnJn-3^2 

^n+1^2g„-l 

m}'n%''i/-' 

85.  How  does  25  x  2  compare  in  value  with  20  x  2  plus 
5x2?  How  is  133,  or  100  +  30+3,  multiplied  by  2  ?  How  is 
the  polynomial  a  -{-b  -'t  c  multiplied  by  the  monomial  m? 

Distributive  Law  for  Multiplication.  —  The  product  of  a  polynomial 
by  a  monomial  is  equal  to  the  algebraic  sum  of  the  partial  prodxicts 
formed  by  multiplying  each  term  of  the  polynomial  by  the  monomial. 

The  Distributive  Law  may  be  established  as  follows : 

Let  a  +  b  he  the  multiplicand  and  m  the  multiplier,  a,  b,  and  m  being 
positive  or  negative  integral  or  fractional  numbers. 

By  the  Law  of  Order  the  multiplier  may  change  places  with  the  multi- 
plicand.    Hence,   (a  +  6)  x  m  may  be  written  m(a  +  6). 

It  is  to  be  proved  that  m(a  +  b)  =  ma  +  mb. 

First,  when  m  is  a  positive  integer. 

Since  m  =  1  +  1  +  1  4-  •••  to  m  terms, 

§  77,    m(a  +  b)  =  (a  +  b)  +  (a  +  b)  +  (a  +  b)+  -•  to  m  terms 
§  56,  z=(a  +  a  +  a  +  ••■to  m  terms)  +  (6  +  6  +  6  +  •••  to  m  terms) 


§77, 


=  ma  +  mb. 


(1) 


Second,  when  m  is  a  fractional  number. 

P 
Let  m  =  —  ,  in  which  p  and  q  are  absolute  integers. 


|(«  +  6) 


by  (1), 


p  times  one  qth  of  (a  +  6) 
=  p(a  +  6)  ^ths 
=  pa  qt\\s,  +-pb  qtha 

=  |ofa  +  |of6 
=  ^a  +  ^b. 

q       q 


(2) 


62  MUL  TIPLICA  TION 

Third,  when  m  is  negative. 

Let  m  =  —  n,  n  being  any  positive  whole  or  fractional  number. 

It  is  to  be  proved  that  (^—  n)(a  +  b)  =  —  na  —  nb. 

By  (1)  and  (2),  n(a  +  b)=na  +  nb.  (3) 

Since,  if  +  n  is  positive,   —  (—  n)  is  also  positive,  substituting  —  (—  n) 

for  +  n  in  (3), 

—  (—  n)(a  +  6)  =  — (—  7i)a  —  (—  n)b 

=  —  (—  va  —  nb).  (4) 

Since  both  —(—  n)(a  +  b)  and   —(—na  —  nb)  are  now  monomial  in 

form,  both  members  of  (4)  may  be  multiplied  by  the  monomial  —  1. 

.'.  +  ( —  n)  (a  +  ft)  =  +  (  —  na  —  nb), 

or  (  —  n)  (a  +  6)  =  —  na  —  nb.  (5) 

By  (1),  (2),  and  (5),        m(a  +  b)=  ma  +  mb 

for  all  positive  or  negative  whole  or  fractional  values  of  m  and  for  all  values 

of  a  and  b. 

86.   To  multiply  a  polynomial  by  a  monomial. 

Examples 
1.   Multiply  Sa^  —  y^  by  —  4  ?/. 

PROCESS  Explanation.  —  By  §  85,  each  term  of  the  multiplicand 

o    ,         2  is  to  be  multiplied  by  the  multiplier. 

The  product  of  .3  x'^  and  —  4  y  is  —  12  x^y.     But  since 
y the  entire  multiplicand  is  3  x^  —  y^,  —  4  ?/  times  ?/2  must  be 


;[2  x^y  -X-  4:ifi    subtracted  from  —  12  x^y.     —  iy  times  y^  =  —  i  y^,  which 

subtracted  from  —  12  x'^y  gives  —  12  x'^y  +  4  y^. 
Or,  since  a  polynomial  multiplied  by  a  monomial  is  equal  to  the  algebraic 
sum  of  the  partial  products  formed  by  multiplying  each  term  of  the  poly- 
nomial by  the  monomial,  §  85,  3  a;^  _  j^2  multiplied  by  —  4  y  is  equal  to 
-  12  x'^y  +  4  2/3. 

Rule.  —  Multiply  each  term  of  the  polynomial  by  the  monomial, 
and  find  the  algebraic  sum  of  the  partial  products. 


2. 

Multiply    2a2-2a6  +  362 
By                                    Sab 

3.                                4. 

5  m^  —  4  71^          ba?  —  2xy  —  y^ 
—  2  m^n                           —  a^y 

Multiply : 

5.  3x2- 2  xy  ^^  53^2^ 

6.  mhi^  —  3  mn*  by  2  mn. 

8.  p^q^  —  2p(f  by  —  pq. 

9.  4a2-5&2c  by  abc^ 

7.   3  a^- 6  a^^  by  -2  6. 

10.  —  2  ac  +  4  aa;  by  —  5  oca?. 

MUL  TI  PLICA  TION  63 

Perform  the  multiplications  indicated : 

1 1 .  a-hc  (3  a*  -  4  a%  -  5  a?b'  +  2  aft^  -  16  6*). 

12.  2xy{o7?-lQxy-^Qy^-5x  +  5y-{-12Qi). 

13.  5  m?  (16  m?  -  20  m^n  + 13  mn^  -  25  «»). 

14.  abc  {a^h''  -  2  a'c'  -  2  6V  -  a*  -  4  5*  -  c*  -  5  ahc). 

15.  -  6c(6^  4- c*  -  &^  -  c^  +  6V  _  4  6^0  +  8  &c2  -  2  he). 

16.  —  2  X  (ic*  —  5  a^y  —  16  0^2/^  +  24 .172/^  —  2/*  —  a:?/  —  a;  +  4). 

87.    To  multiply  a  polynomial  by  a  polynomial. 

To  multiply  p  +  q  +  r  by  a  +  6, 

§  85,  {p  +  q-{-r){a  +  h)=p{a  +  h)  +  q{a  +  h)+r{a  +  h) 

§§82,85,  =ap  +  hp  +  aq  +  hq-\-ar  +  hr 

§  55,  =  ap  +  aq  -\-  ar  +  hp  +  hq  +  br. 

Rule.  —  Multiply  every  term  of  the  multiplicand  by  each  term  of 
the  multiplier,  and  find  the  algebraic  sum  of  the  partial  products. 

Examples 
1.    Multiply  x^  —  xy  by  2  a;  +  3  ?/. 

PROCESS 

a?  —  oey 
2x  +3?/ 


+  2  a;  times  {a^  —  xy)  =  2  a^  —  2  a^y 

-\-Sy  times  (ar^  —  xy)  =  +  3  a^y  —  3  xy^ 

.-.  (2  X  +  3  y)  times  (a^  —  xy)  =  2a^4-    x^y  —  3xy^ 

2.    Multiply  x^  —  3x^y  +  3xy^  —  y^  by  t?  —  xy. 

PROCESS 

3?-3x'y  +  3xy''  -f 
ai?  —  xy 


af-3ai*y  +  3a^f-    x'f 
—    x*y  -\-  3  x^y^  —  3  a^y'  +  xy* 

0^—4  a;*2/  +  6  x^y^  —  4  a^?/^  +  xy* 


64 


MUL  TIPLICA  TION 


3.    Multiply  Sx  +  af'  —  2aj2  —  X*  by  Sx  +  l  —  a^. 


PROCESS 

5x—    2a^  -{-  Jt^  —  X* 
1     +    3a;  -x" 
5x-    2 
15 


C»2+     1 

3^-1 

x* 

-    6 

+  3 

-3 

-    5 

+  2 

-1 

x* 


+  «« 


6  a;  +  13  x^  -  10  x^  +  4  a;*  -  4  af  +  x" 


TEST 

=  +  3 

=  +  3 


=  +  9 


Suggestion.  —  For  convenience  in  writing  partial  products,  both  polyno- 
mials are  arranged  so  that  in  passing  from  left  to  right  the  several  powers  of 
X  are  either  successively  higher  or  lower.  In  this  process,  the  polynomials 
are  arranged  according  to  the  ascending  powers  of  x. 

Test.  —  Since  the  correct  product  of  5  x  —  2  x^  +  x'  —  a:*  and  1  +  3  x  —  x^ 
is  the  same  in  form  whatever  value  x  represents,  it  is  possible,  by  assigning 
an  arithmetical  value  to  x,  to  change  the  process  of  multiplying  one  algebraic 
expression  by  the  other  into  a  process  of  multiplying  one  arithmetical  num- 
ber by  another  as  shown  in  the  test. 

Let  1  be  substituted  for  x. 

Multiplicand  =  5x-2x2  +  x3-x<  =  5-2-fl-l=  +  3 

Multiplier      =  1  +  3  x  -  xJ^ =  1  -f- 3  -  1         =4-3 

Product  should  be  equal  to  4-9 

5  X  4- 13  x2  -  10  x3  +  4  X*  -  4  x5  4-  x6  =  5  4- 13  -  10  4-  4  -  4  4-  1  =  4-  9. 

In  like  manner  the  multiplication  of  any  two  literal  expressions  may  be 
tested  arithmetically  by  assigning  any  values  we  please  to  the  letters.  It  is 
usually  most  convenient  to  substitute  4-  1  for  each  letter,  since  this  may  be 
readily  done  by  adding  the  numerical  coefficients. 

Multiply,  and  test  each  result : 
4.    2a;H-3  by  x-f-2.  12.    Ay —  6b  hy  2y-\-b. 


5.  4x-|-l  by  3x-|-4. 

6.  5  71  —  1  by  4  ?i  -1-  5. 

7.  /i  +  2fc  by  3h-k. 

8.  3r-6sby5r  —  2s. 

9.  4r-h2sby2r-f-9s. 

10.  3l  +  5t  by  2l  +  6t. 

11.  4  a -I- 3  X  by  4  a  — 3  X. 


13.  Sx  —  2y  by  3x  +  2y. 

14.  26  +  5c  by  5&-2c. 

15.  7x  — 2«  by  4x-|-2w. 

16.  ab  —  15  by  a& -f  10. 

17.  ax -{-by  by  ax  — by. 

18.  a^  —  ay  +  y^  by  a  +  y. 

19.  3a^-Qab+3b^  by  2a-3  6. 


M  UL  TI PLICA  TION  65 

Multiply : 

20.  2a?-ZW-ab  by  3  a='-4ct6-5  &'. 

21.  5a;-5.T2  +  10  by  12-^0x  +  2x\ 

22.  3n^  +  3m2  +  mw  by  m^  — 2mn2  +  m^n. 

23.  4y2-10  +  2y  by  22/^-32/  +  5. 

24.  4  a;  -  3  a;2  +  2  a^  by  3  a;  - 10  a^  + 10. 

25.  a^  +  a*  +  4  a^  -  a^  +  a  by  a  + 1- 

26.  31a^-27a^  +  9x-3  by  3a;  +  l. 

27.  4  0^  -  3  ic^^  +  5  a;^^  -  6  2/3  by  5  tc  +  6  y. 

28.  a-\-h  +  c-\-dhja-h  —  c  +  d. 

29.  a2  +  62_^c2-a&  — ac  — 6c  by  a  +  &  +  c. 

30.  aa^"  +  a2/^"  by  a^^'  —  af''. 

31.  aa;"-i  +  2/"~'  by  3  aaf-^  +  2  .v"-^ 

32.  ar^"  +  2  a;"^"  +  2/'"  by  ar'"  -  2  a;»t/"  +  2/'"- 

An  indicated  product  is  said  to  be  expanded  when  the  multipli- 
cation is  performed. 

Expand : 

33.  (x  +  y){x  +  y).  39.   (2a2  +  6)(2a2-6). 

34.  (a +  &)(«  +  &).  40-  ('K" +  ?/")(«'" -2/"> 

35.  (c3+d3)(c-^  +  d3).  41.  (3aa:  +  2  6y)(3ax  +  2  62/). 

36.  (a;»  +  r)  (^"  +  2/")-  *2.  (3  aa;  +  2  &?/)  (3  aa;  -  2  &^). 

37.  (3  a +  6) (3 a +  6).  43.  (4m- 5n)(4m  +  5w). 

38.  (3a  +  &)(3a-6).  44.  (a  +  &  +  c)(a  +  6-c). 

45.  (a  +  6)(a-6)(a  +  6)(a-6). 

46.  {a'  +  x^{a'-^)(a'  +  x')(a'-x^. 

47.  (a-&)(a-f6)(a2  +  6^(a*  +  6*). 

48.  (a"  -  6")  (a"  +  6")  («='"'  +  &'")• 

49.  (2a  +  36  +  5c)(2a  +  36-5c). 

50.  (5  m  -  2  rt  +  a;)  (5  m  -  2  w  -  a;). 

51 .  (a;»  -  wa;"--Vy  +  I  nx"-y)  (x  +  y). 

52.  (a;"  +  3  ajp-^z  -  6  xP- V)  (ar' -  ?/^. 


66  MULTIPLICATION 

88.  When  polynomials  are  arranged  according  to  the  ascending 
or  the  descending  powers  of  some  literal  factor,  processes  may 
frequently  be  abridged  by  using  the  Detached  Coefficients. 

53.   Expand  (2a;*-3a^  +  3a;  +  l)(3a;  +  2). 

FULL    PROCESS  DETACHED    COEPPICIENTS 

2  a;^-3  0^+3x4-1  2-3+0  +3  +1 

3  a; +2  3+2 


6a^-9.'B*    +9i»2+3a;        6  -9  +0  +9  +3 

4a;*-6a^     +6a;+2  4  -6  +0  +6  +2 

6  af-5  x*-6  a^+9  x'+'d  x+2  6  a^-5  a;*-6  a^+9  a^+9  x+2 

Since  the  second  power  of  x  is  wanting  in  the  first  factor,  the  term,  if 
it  were  supplied,  would  be  Ox',  and  the  detached  coefficient  of  the  term 
would  be  0. 

The  detached  coefficients  of  missing  terms  should  be  supplied  to  prevent 
confusion  in  placing  the  coefficients  in  the  partial  products  and  to  prevent 
errors  in  determining  the  result. 

54.   Multiply  a*  -  2  a^ft  +  3  a^b^  -5ab^  +  5b'  by  a  +  2  &. 

PROCESS  TEST 

l_2+3-5+   5  =2 

1  +  2  =3 


1-2+3-5+   5 

2-4  +  6-10  +  10 

1-1-0-1+1-5  +  10  =6 

=  a'  +  Oa*b-  a?b^  +  a?W  -  5  a6*  + 10  6* 
=  a^  -  a?l^  +  a?¥  -  5  a6*  + 10  b\ 

Observe  that  the  powers  of  a  decrease  uniformly  from  left  to  right,  and 
that  the  powers  of  b  increase  uniformly  from  left  to  right. 

Observe  also  that  the  sura  of  the  exponents  is  the  same  in  every  term. 

Expand  the  following,  using  detached  coefficients,  and  test  the 
results : 

55.  {ix^-\-4:^y -\-Qx^y^ -^-^LXif -\-y*){x->ry). 

56.  {5a?-x' -2a^  +  x'  +  ^:i?-l){x  +  2). 


MULTIPLICATION  67 

57.  (a'  +  4a='+16a-32)(a3  +  a2  +  a-Hl). 

58.  {p"  -  2pq  +  q")  (p'  +  2pq  +  q^. 

59.  (a;-l)(a;-2)(a;-3)(a;  +  4)(.'K  +  2). 

60.  (15 r*s  -10rs*  +  7^  +  ^  +  3?V  +  Si^sP)(r  -2s). 

61.  (x^^  -  a^  +  cc^  -  x'  -{-  x^  -  x^  -^  X*  -  a^  +  a^  -  X  +  l)(x  +  1). 

62.  (x^o  +  3  x»  -  2  a*  +  5  a;*  +  2  ar^  +  2)  (cB«  +  «*  +  ar'  +  1). 

63.  (a^  +  a:^  _,_  ^,2  ^  a;  ^  j^>)  (^^4  _  3^  _l_  a^  _  g,  ^  1^  (^a;  ^  1^ (a;  - 1). 

64.  (x^  —  4:2i^y  +  5 afy^  +  3 a^y*  —  x'f  +  xf  -  /)(2  a;  —  S^/). 

Expand : 

65.  (2a^  +  4.'B2  +  8a;  +  16)(3a;-6). 

66.  (a^  +  4:0^ +  5x-24:)(a^- 4.x +  11). 

67.  (a^-4a:2^11a;-24)(ar'  +  4a;  +  5). 

68.  (a^  -  a;  4- 1)  (ar'  +  a;  +  1)  (a;*  —  a^  +  1). 

69.  (16a*-8a''  +  4a2-2a  +  l)(2a  +  l). 

70.  (a^  -  2  a^  +  3  a  -  4)  (4  a^  +  3  a^  +  2  a  +  1). 

71 .  (m*  +  2m^-\-m''-4m-  11) (m=^ - 2 m  +  3). 

72.  (m*  +  2  m^n  +  2  mn-  +  w^)  (^n,^  —  2  m'^w  +  2  mn'  —  n*). 

73.  (a:3-La^  +  ^a;-T'^)(a;  +  i). 

74.  (^x3  +  ^^  +  |^4.2)(a;-|). 

76.  fa^-^Vsa'  +  2aV3a-3\ 

77.  (a;"-^  —  a;"-2  +  a;»-3  —  a;"-*)  (a;  +  1). 

78.  (1  -  a"  +  or"  -  a^")  (1  -  a"). 

79.  (a^"  +  2  a"6"  +  ft^™^  (^^r,  _  j„^_ 

80.  (ar^"  —  x^j/™  +  2/^")  (a;"  +  2/"). 


58  MULTIPLICATION 

Equations  and  Problems 
89.   1.  Given  5  (2  a; -3)  =7(3  a; +  5) -72,  to  find  the  value  of  x. 

Solution 
5(2x-3)=7(3x  +  6)-72. 
Expanding,  10  «  -  15  =  21  x  +  35  -  72. 

Transposing,  10  x  -  21  x  =  15  +  35  -  72. 

Uniting  terms,  —  11  x  =  —  22. 

Multiplying  by  -  1,  11  x  =  22. 

.-.  x  =  2. 
Verification.  —  Substituting  2  for  x  in  the  given  equation, 
5(4 -3)=  7(6 +  5) -72. 
5  =  77  -  72  =  5. 

Find  the  value  of  x,  and  verify  the  result,  in  the  following : 

2.  4  =  5-(x-2)-(x-3).         5.   10iK-2(a;-3)  =  -10. 

3.  2  =  2a; -5 -(x -3).  6.   6a;-3(a;-6)=4(2a:-l)  +  2. 

4.  1  =5(2a;-4)  +  5.T-f-6.  7.   7(5  -  3a;)  =  3(3  -  4a;)  -  t 

8.  4(2-4a;)  =  4-2(a;  +  5). 

9.  5  +  3a;-4  =  13  +  4(a;-4). 

10.  49-2(2a;  +  3)  =  9  +  2(2a;-3). 

11.  3a;-2(4a;-5)  =  -2(6  +  2a;). 

12.  3(a;  +  l)-2(2a;  +  5)  =  6(3-a;). 

13.  2(a;-5)  +  7  =  a;  +  30-9(a;-3). 

14.  5  +  7(a;-5)  =  15(a;  +  2-36). 

15.  (a; -2)  (a; -2)  =  (a; -3)  (a; -3) +  7. 

16.  (a;-4)(a;  +  4)  =  (a;-6)(a;  +  5)  +  25. 

17.  4ar'-4(a.-3-a;2_^3,_2)  =  4a^. 

18.  7(2a;-3  6)  =  26-3(2a;  +  &). 

19.  lla=3(.T-2a)-5(2a;-2a). 

20.  3(2&-4a;)-(a;-6)=-6  6.     24.    3  (a;-a-2  6)=3  ft. 

21.  4a;-x-2=a;(2— a;)+2a.  25.    5  ft=3  (2  a;— 6)— 4  6. 

22.  2(a;  +  d)=10c.  26.    13  (a:-a)=5  (2 a;+a). 

23.  5c=b{x+a-b).  27.    5  (4  a;-3  a)-6  (3  a;-2  a)=3  a. 


I 


MULTIPLICATION  59 


Solve  tlie  following  problems  and  verify  the  solutions : 

28.  George  and  Henry  together  had  46  cents.     If  George  had 

4  cents  more  than  half  as  many  as  Henry,  how  many  cents  had 

each? 

First  Solution 

Let  X  =  the  number  of  cents  George  had. 

Then,  «  —  4  =  the  number  of  cents  George  had  less  4, 

and  2  (x  —  4)  =  the  number  of  cents  Henry  had  ; 

/.  X  +  2  (a:  -  4)  =  46. 

Expanding,  x  +  2x  —  8  =  46; 

.-.  X  =  18,  the  number  of  cents  George  had, 

and  2  (x  —  4)  =  28,  the  number  of  cents  Heniy  had. 

Second  Solution 
Since  George  had  4  cents  more  than  half  as  many  as  Henry, 
let  2  X  =  the  number  of  cents  Henry  had ; 

then,  X  4-  4  =  the  number  of  cents  George  had, 

and  2  X  +  X  +  4  =  46  ; 

.-.  X  =  14, 

2  X  =  28,  the  number  of  cents  Henry  had, 
and  X  +  4  =  18,  the  number  of  cents  George  had. 

Verification 

The  answers  obtained  should  be  tested  by  the  conditions  of  the  problem. 
If  they  satisfy  the  conditions  of  the  problem,  the  solution  is  presumably 
correct. 

1st  condition  :  —  They  had  together  46  cents. 

18  +  28  =  46. 
2d  condition  :  —  George  had  4  cents  more  than  half  as  many  as  Henry. 
18  =  I  of  28  +  4. 

29.  In  a  certain  election  at  which  8000  votes  were  polled,  B 
received  500  votes  more  than  half  as  many  as  A.  How  many 
votes  did  each  receive  ? 

30.  A  had  $40  more  than  B;  B  had  f  10  more  than  one  third 
as  much  as  A,     How  much  money  had  each  ? 


60  MULTIPLICATION 

31.  Mary  is  25  years  younger  than  her  mother.  If  she  were 
one  year  older,  she  would  be  \  as  old  as  her  mother.  What  is 
the  age  of  each  ? 

32.  If  John  had  3  more  marbles,  he  would  have  3  times  as 
many  as  Clarence.  Both  have  41  marbles.  How  many  has 
each? 

33.  Two  boys  together  had  $8.20,  and  one  had  50  cents  less 
than  half  as  much  as  the  other.     What  amount  had  each  ? 

34.  If  5  is  subtracted  from  twice  a  certain  number,  and  the 
difference  is  multiplied  by  3,  the  result  is  9  less  than  5  times  the 
number.     What  is  the  number  ? 

35.  A  is  f  as  old  as  B;  8  years  ago  he  was  ^  as  old  as  B. 
What  is  the  age  of  each  ? 

Suggestion.  —  Let  5  x  =  the  number  of  years  in  B's  present  age. 

36.  In  2  years  A  will  be  twice  as  old  as  he  was  2  years  ago. 
How  old  is  he  ? 

37.  Two  wheelmen  start  at  the  same  time  from  A  to  ride  to  B. 
One  rides  at  the  rate  of  10  miles  an  hour,  and  rests  3  hours ;  the 
other  rides  at  the  rate  of  8  miles  an  hour,  and  by  resting  only 
1  hour  arrives  at  B  as  soon  as  the  faster  rider.  How  far  is  it 
from  A  to  B,  and  how  many  hours  are  occupied  in  making  the 
trip  ? 

38.  A  man  had  two  flocks  of  sheep  with  the  same  number «f 
sheep  in  each.  After  selling  100  sheep  from  one  flock,  and  20 
from  the  other,  the  numbers  remaining  were  as  2  to  3.  How 
many  sheep  had  he  in  each  flock  at  first  ? 

39.  Mary  bought  17  apples  for  61  cents.  For  a  certain  num- 
ber of  them  she  paid  5  cents  each,  and  for  the  rest  she  paid  3 
cents  each.     How  many  of  each  kind  did  she  buy  ? 

40.  George  is  \  as  old  as  his  father ;  a  years  ago  he  was  \  as 
old  as  his  father.     What  is  the  age  of  each  ? 

41.  A  rug  is  3  feet  longer  than  it  is  wide.  When  it  is  placed 
on  the  floor  of  a  certain  room,  it  leaves  a  margin  of  2  feet  on 
every  side.  If  the  area  of  the  floor  is  172  square  feet  greater 
than  that  of  the  rug,  what  are  the  dimensions  of  the  floor  ? 


MULTTPLICA  TION 


61 


SPECIAL  CASES  IN  MULTIPLICATION 

90.  The  square  of  the  sum  of  two  numbers. 

(a  +  h){a  +  h)=a?  +  2ah  +  h\ 
{x  +  y){x  -^y)=x'  +  2xy  +  f. 

1.  When  a  number  is  multiplied  by  itself,  what  power  is 
obtained  ?    What  is  the  second  power,  or  square  of  (a  +  6)  ? 

of  (.x-  +  t/)? 

2.  How  are  the  terms  of  the  square  of  the  sum  of  two  numbers 
obtained  from  the  numbers  ? 

3.  What  signs  have  the  terms  ? 

91.  Principle.  —  The  square  of  the  sum  of  two  numbers  is  equal 
to  the  square  of  the  first  number,  plus  twice  the  product  of  the  first 
and  second,  plus  the  square  of  the  second. 

Since  5  a^  x  5  a^  =  25  a«,  3  a*b^  x  3  a*b'  =  9  a^ft^",  etc.,  it  is  evident 
that  in  squaring  a  number  the  exponents  of  literal  factors  are 
doubled. 

Examples 

Expand  by  inspection : 


1. 

(m  +  n){m  +  n). 

13. 

(3b  +  cy. 

2. 

(P  +  <i)(P  +  Q)' 

14. 

(2a +  3  6)2. 

3. 

(r  +  s){r  +  s). 

15. 

{Bx  +  2yy. 

4. 

(a  +  x){a  +  x). 

16. 

(72  +  3  cf. 

5. 

{x -\- 4:){X  +  A). 

17. 

(3  6  +  10  xf. 

6. 

(m  4-  5)(m  +  5). 

18. 

(a^  +  by. 

7. 

(a  +  6)(a  +  6). 

19. 

(a'  +  by. 

8. 

(2/  +  7)(2/  +  7). 

20. 

(a»  +  6")'. 

9. 

(z+l){z  +  l). 

21. 

(x^+y")'. 

10. 

(2  a  +  x)(2  a  +  x). 

22. 

(3  a* +  5  6^1 

11. 

(3  m  -1-  n)(3  m  +  n). 

23. 

(1  +  2  a'bY. 

12. 

{5  x  +  z){o  X  +  z). 

24. 

(^-i  +  y-iy. 

62  MUL  TIP  Lie  A  TION 

25.    Find  the  square  of  41. 

Solution 
412  =  (40  +  1)^  =  402  +  2  X  40  X  1  +  12  =  1681. 


Square : 

26.   21. 

29. 

45. 

32. 

22. 

35. 

81. 

27.    24. 

30. 

83. 

33. 

72. 

36. 

91, 

28.    25. 

31. 

65. 

34. 

43. 

37. 

101, 

38.    Find  the  square  of  1\. 

•  Solution 

(7J)2=(7  + 1)2  =  72  +  2  X  7x  i+(l)2  =  49+7  +  J  =  56|. 

Observe  that  the  middle  term  of  the  square  of  any  number  expressed  by  an 
integer  and  the  fraction  \  is  equal  to  the  integer.  Hence,  the  square  of  such 
a  number  is  equal  to  the  square  of  the  integer,  +  tlie  integer,  +  tiie  square 
of  the  fraction.  Observe  also  that  the  sum  of  the  first  two  terms  of  the  square 
may  be  found  by  multiplying  the  integer  by  tlie  integer  increased  by  1. 


Thus, 

(3i)2  =  9  +  3  +  i  =  12i, 

or 

(3^)2  =  3  X  4  +  i  =  12i. 

Find  the  square  of 

39.    5^.                 41. 

2f                  43.    1.5. 

45.  6.5. 

40.    41                   42. 

12^.                44.    5.5. 

46.  8.6. 

92.  The  square  of  the  difference  of  two  numbers. 

(a  -  h){a  -b)=a'  -2ab-\-  h\ 
{x  -  y){x  -y)=;t?-2xy  -\-  y\ 

1.  What  is  the  square  of  (a  —  h)  ?   of  {x  —  y)  ? 

2.  How  is  the  square  of  the  difference  of  two  numbers  obtained 
from  the  numbers  ? 

3.  How  does  the  square  of  (a  —  h)  differ  from  the  square  of 
(a+6)? 

93.  Principle.  —  The  square  of  the  difference  of  two  numbers  is 
equal  to  the  square  of  the  first  mimber,  minus  twice  the  product  of 
the  first  and  second,  lilus  the  square  of  the  second. 


\ 


MULTIPLICA  TION 

68 

Examples 

i)xpa 

ind  by  inspection: 

1. 

[x  —  m)  (a-  —  m). 

10. 

(2  a-  x)\ 

19. 

(Sx-2y. 

2. 

(m  —  n)  (m  —  n). 

11. 

(3  m-  n)\ 

20. 

(2x-5yy. 

3. 

(x  _  6)  (a;  -  6). 

12. 

{4x-yy. 

21. 

(2x-4.yy. 

4. 

(i>-8)(p-8). 

13. 

(5  m  —  ny. 

22. 

(5  m  —  3  ny. 

5. 

(5 -7)  (9 -7). 

14. 

(m  —  4  n)^ 

23. 

i3p-oqy. 

6. 

(a  —  c)  (a  —  c). 

15. 

(p-3qy. 

24. 

(a"  _  6")2. 

7. 

(^•-0(»--0- 

16. 

(a  -  7  6)2. 

25. 

{x^-yy. 

8. 

'a  —  x)  (a  —  cc). 

17. 

(4  a -3)2. 

26. 

(a--x_2/»-i)«. 

9. 

{:x-l){x-l). 

18. 

(5ic-4)2. 

27. 

(ma;™  —  wy")*. 

28.  Find  the  square  of  19. 

Solution 
192  =  (20  -  1)2  =  202  -  2  X  20  X  1  +  12  =  361. 

Find  the  square  of 

29.  49.  32.  29.  35.  67. 

30.  69.  33.  38.  36.  89. 

31.  79.  34.  48.  37.  99. 


38.  998. 

39.  999. 

40.  595. 


94.    The  square  of  any  polynomial. 

By  actual  multiplication, 

(a+b  +  cy=a^+b^+c'-\-2ab-\-2ac+2bc. 

(a+b-c+dy'=a^+b--\-cr+cl^+2ab-2ac+2ad-2bc-]-2bd-2cd. 

(a-\ \-k-\ \-my=a^-^ \-J(^-\ [-m^  +  2ak-\ 

+  2amA \-2km-i . 

1.  In  the  square  of  each  polynomial  what  terms  are  squares? 

2.  How  are  the  other  terms  formed  from  the  terms  of  the 
polynomial  ? 

3.  What  signs  have  the  squares  ?     How  are  the  signs  of  the 
other  terms  determined  ? 


64 


MUL  TIPLICA  TION 


95.  Principle.  —  The  square  of  a  polynomial  is  equal  to  the 
smn  of  the  squares  of  the  terms  and  tioice  the  product  of  each  term 
by  ea^h  term  that  follows  it. 

Examples 

Expand  by  inspection : 

4.  (x-y  +  zf.  7.    (a-2b  +  cy. 

5.  (^x  +  y-3zy.  8.    (2a-b-cY. 

6.  (x-y  +  3zy.  9.    (b-2a  +  cy. 

15.  (2a-5b  +  3cy. 

16.  (2  m  — 4:  n  —  ry. 

17.  (12-2x  +  Syy. 

18.  (a-\-m  +  b  +  ny. 

19.  (a  —  m  +  b  —  ny. 

96.  The  product  of  the  sum  and  difference  of  two  numbers. 

(a  +  6)  (a  -  6)  =  a2  _  ^2^ 
(x"  +  y")  (x"  —  ?/")  =  a^"  —  /". 

1.  Since  a  +  6  represents  the  sum  and  a  —  b  the  difference  of 
two  numbers,  to  what  is  the  product  of  the  sum  and  the  difference 
of  two  numbers  equal  ? 

2.  How  are  the  terms  of  the  product  obtained  from  the 
numbers  ? 

3.  What  sign  connects  the  squares  ? 

97.  Principle.  —  TJie  product  of  the  sum  and  difference  of  two 
numbers  is  equal  to  the  difference  of  their  squares. 


1. 

[x  +  y  +  zy. 

2. 

[x  +  y-zy. 

3. 

[x-y-  zy. 

10. 

[ax  —  by  +  czy. 

11. 

[ma  +  n6  —  rzy. 

12. 

[qb  —  pc  —  rdy. 

13. 

[ac  —  bd  —  dey. 

14. 

[3x-2y  +  4z) 

Examples 


Expand  by  inspection ; 

1.  (x  +  y)(x-y). 

2.  (a  +  c)(a—  c). 

3.  {m  +  n)  (m  —  n). 

4.  (p  +  9)(i>-^). 

5.  (2>  +  5)(p-5). 


6.  (r  +  s)  (r  —  s). 

7.  (a;  +  1)  (a;  -  1). 

8.  {:)^  +  l){;t^-l). 

9.  {^+l){o?  -1). 
10.  (a;*  -  1)  (»*  + 1). 


MUL  TIPL ICA  TION  65 

11.  (ar>  -  1)  (ar* -f- 1).  20.    {2a^  +  5y')(2a^  -  5f). 

12.  (x^  +  f)(x^^-f).  21.    {3a^  +  2y')(Sx'-2f). 

13.  {ab  +  5){ab-5).  22.    {20^ -h2b-'){2a' -2  b'). 

14.  (cd  +  d^)  (cd  -  d^).  23.    (-5n-26)(-5n  +  26). 

15.  (2a;+32/)(2a;-3y).  24.    (- cc- 2y)(- a;  +  2?/). 

16.  (3m  +  4n)(3OT-4w).        25.    (- 5  -  3m)(- 5  +  3m). 

17.  (12-\-xy){12-xy).  26.    (ic'»-^  +  ^»+i)(a;'»-i  -  ?/»+^). 

18.  (3  m^w  -  6)  (3  m^'n  +  6).        27.    (3  x"  +  7  y")  (3  x™  -  7  y"). 

19.  (a&  +  cd)  (a6  -  cd).  28.   (5  a^b^  +  2  x*)  {5  a^b^  -  2  x*). 
One  or  both  of  the  numbers  may  consist  of  more  than  one  term. 

29.  Expand  (a  +  m  —  w)  (a  —  m  +  n). 

Solution 

a  +  m  —  n  =  a  +  (m  —  n). 

a  —  m  +  n  =  a  —  (m  —  n). 

.'.  [_a-{-m  —  n}la  —  »»  +  n]  =  [a  +  (m  —  n)'\la  —  (m  —  n)]. 

Prin.,  =  a^  —  (m  —  n)^ 

§  93,  =  a2  -  (m2  -  2  mn  +  n«) 

=  a2  -  m2  4-  2  mn  -  n^. 
Expand : 

30.  (a-\-x-y){a-x-]-y).  36.  (y  +  c -{-d){y  +  c -d). 

31.  (x  +  c  — d)(iB  — c +  d).  37.  (a-\-x  + y){a  +  x—y). 

32.  (r+i3-9')(r-i3  +  ^).  38.  (o^  +  2x  +  l)(a^  +  2a; -  1). 

33.  (r  +  p  +  q)(r-p-q).  39.  (ar*  +  2  x  -  1)  (x^  _  ^  a,  ^  ^-^ 

34.  (x  +  b  +  n)(x-b-n).  40.  (x^^Sa;- 2)(r' -  3a;  + 2). 

35.  (2/  +  c  +  d)(2/-c-d).  41.  (a;2  +  3  a;  +  2)  (ar' -  3  a;  +  2). 

42.  (m*-2m'  +  l){m*  +  2m'  +  l). 

43.  (2x  +  Sy-4:z){2x-\-3y  +  4:Z). 

44.  (2x*-a;2/  +  32/2)(2ar'  +  xi/-3  2r^. 

45.  (ar*  +  ajy  +  2/^)  (ar*  -  a^  +  2/^. 

ALG.  — 5 


66 


MULTIPLICATION 


46.  [(a  +  &)+(c-f  d)][(a  +  6)-(c  +  d)]. 

47.  {a-{-h  +  x  +  y){a-\-h —  x  —  y). 

48.  {a  +  h  +  m  —  n){a  -^  b  —  m  -\-  n). 

49.  (x  —  m  -{-  y  —  n)  (x  —  m  —  y  -\-  n). 

50.  {p  —  q  -\-r-\-  s){p  —  q  —  r  —  s). 

51.  (a  —  m  —  6  —  w)  (a  +  m  —  6  +  w). 

52.  {a-\-x  +  h  ~y){a~x  +  b -{-y). 

53.  Find  the  product  of  32  x  28. 

Solution 
32  X  28  =(30 +  2)  (30 -2) 

=  302  -  22  =  900  -  4  =  896. 

Find  the  product  of 

54.  31  X  29.  57.  48  X  52.  60.  98  x  102. 

55.  42  X  38.  58.  57  X  63.  61.  99  x  101. 

56.  69  X  71.  59.  95  X  85.  62.  95  x  105. 

63.  What  is  the  square  of  95  ? 

Solution 
(a  +  6)(a-6)=a2_52.  ^Ij 

Transposing,  etc.,  a'^  =  (a  +  b)  (a  —  6)  +  6^.  f2) 

Let  a  =  95  and  6  =  5. 

Equation  (2)  becomes         952  _  (95  +  5)  (95  _  5)  +  52 

=  100  X  90  +  25  =  9025 

64.  What  is  the  square  of  48  ? 

Solution 
Let  a  =  48  and  6  =  2. 

Equation  (2)  becomes        482  _  (43  +  2)  (48  -  2)  +  2* 

=  50  X  46  +  4  =  2304. 

Square  by  a  similar  process : 

65.  98.  67.   93.  69.   58.  71.   87.  73.     68. 

66.  96.  68.    97.  70.    49.  72.   79.  74.    129. 


MUL  TIPLICA  TION  67 

98.  The  product  of  two  binomials  that  have  a  common  term. 

{x  4-  2){x  +  5)=  af  +  2  X  -1-  5  a;  4- 10 

^x'  +  l  x  +  10. 
{x  +  2){x  -  5)=  x-  +  2  X  -  5  a;  -  10 

=  af'  -  3  a;  -  10. 
{x  -  2){x  -b)=x'-2x-hx  +  10 

=  a.-^  -  7  a:  +  10. 
(x  +  a)(x  —  6)  =  .x*^  -f  ax  —  bx  —  ab 

=  XT  -\-{a  —  b)x  —  ab. 

1.  How  many  terms  are  alike  in  each  pair  of  factors  ? 

2.  How  is  the  first  term  of  each  product  obtained  from  the 
binomial  factors  ? 

3.  How  is  the  third  term  of  each  product  obtained  from  the 
factors  ? 

4.  How  is  the  second  term  of  the  product  in  the  first  example 
obtained  from  the  factors  ?  in  the  second  example  ?  in  the  third 
example  ?  in  the  fourth  example  ? 

5.  How  can  the  formation  of  the  second,  or  middle  term  be 
described  so  as  to  apply  to  all  of  the  examples  ? 

99.  Principle.  —  Tlie  product  of  two  binomials  that  have  a  com- 
mon term  is  the  algebraic  sum  of  the  square  of  the  common  term,  the 
common  term  multiplied  by  the  algebraic  sum  of  the  unlike  terms, 
and  the  algebraic  product  of  the  unlike  terms. 

Examples 
Expand  by  inspection : 

1.    (x  +  5)(x-^6).  7.    (a;  -  5)(a;  -  1). 


2.  (a;  +  7) (a;  4- 8) 

3.  (a;-7)(a;  +  8) 

4.  (a;  +  7)(a;-8) 

5.  (x  -  5)(x  —  4) 

6.  (a;-3)(a;-2) 


8.  (x  +  5)(x  +  8). 

9.  (p-4)(p  +  l). 

10.  (r-3)(r-l). 

11.  (m-6)(m  +  5). 

12.  (m  -  2)(m  4- 10). 


68 


M  UL  TIP  Lie  A  TION 


13. 

(n-8)(ri  -12). 

24. 

(jj-2a){y  +  ^h). 

14. 

(ii-6)(?6  +  15). 

25. 

(2-4  a)  (2 +  3  a). 

15. 

(ar'  +  5)(ar-3). 

26. 

(2  a; +  5)  (2  a; +  3). 

16. 

(a^-7)(ar'  +  6). 

27. 

(2  a; -7)  (2  a; +  5). 

17. 

(x-*-3)(ir*  +  9). 

28. 

(3.v-l)(32/  +  2). 

18. 

(x  +  c){x  +  d). 

29. 

(4a^+l)(4ar^-7). 

19. 

(m  +  a)  (m  +  6). 

30. 

(a&-6)(a6+4). 

20. 

(r  +  a)(r-6). 

31. 

(ar^/-a)(a^y^  +  2a). 

21. 

(s  —  a)  (s  +  w). 

« 

32. 

(2  m  -  ab)  (2  m  +  3  a6). 

22. 

(a;"  -  5)  (x"  +  4). 

33. 

(5i?-ac2)(2ao2  +  5i>). 

23. 

(a;"  -  a)  (x^  -  b). 

34. 

(3xy-\-9/){y^-xy). 

35.   (a 

+  b  +  5){a 

+  6  +  2). 

36.    (a 

-h- 

-4)(o 

i-h 

+  10). 

37.    (x 

+  y- 

-1)(^ 

+  2/  +  2). 

38.  (x 

39.  (^ 

-y- 

-2){x 

-i)C 

-y 

-8). 

+  x 

^  + 

x  +  3). 

40.    (2  m  +  w  -  3)  (2  m  +  »  +  4). 

By  an  extension  of   the  method  given  above,  the  product  of 
any  two  binomials  may  be  written. 

41.   Expand  (3x +  2y)(5x  —  4:y). 

Solution 
(3x  +  2y)(5x-^y)-  15x2+  \Oxy  -  \2zy-8y' 
=  15  x2  -  2  aiy  -  8  y^. 


Expand  by  inspection : 

42.  (2x  +  5y)(3x-\-4.y). 

43.  (3x-4:y)(2x  +  5y). 

44.  (3a-66)(2a  +  3&). 


45.  (3  a2  _  1)  (2  a^  +  3). 

46.  (m^  +  n)  {2  m^  -  n). 

47.  (a  +  26)(c-2d). 


MULTIPLICATION  69 


Exercises 


100.  1.  In  a  certain  family  there  are  three  children  each  of 
whom  is  2  years  older  than  the  one  next  younger.  When  the 
youngest  is  x  years  old,  what  are  the  ages  of  the  others  ?  When 
the  oldest  is  y  years  old,  what  are  the  ages  of  the  others  ? 

2.  What  two  whole  numbers  are  nearest  to  the  whole  num- 
ber X? 

3.  Mary  read  10  pages  in  a  book,  stopping  at  the  top  of  page  a. 
On  what  page  did  she  begin  to  read  ? 

4.  A  man  made  three  purchases  of  a,  b,  and  2  dollars,  respec- 
tively, and  tendered  a  10-dollar  bill.  Express  the  number  of  dol- 
lars change  due  him. 

5.  A  sold  B  grain,  hay,  and  potatoes  for  a,  h,  and  c  dollars, 
respectively ;  but  some  of  the  grain  becoming  damaged,  and 
some  of  the  potatoes  having  been  frozen,  he  deducted  x  +  y 
dollars  from  B's  indebtedness.  If  B  offered  a  100-dollar  bill  in 
payment,  what  was  the  amount  due  him  in  return  ? 

6.  What  is  the  cost  of  5  apples  at  b  cents  each?  What  will 
a  apples  cost  at  b  cents  each  ? 

7.  How  many  cents  are  there  in  a  dollars  ?  How  many 
dimes  are  there  in  b  dollars  ?   in  ax  dollars  ? 

8.  If  a  man  earns  $2  per  day,  how  much  will  he  earn  in 
a  days  ?  in  c  days  ?   in  a  —  c  days  ? 

9.  How  much  will  a  man  whose  wages  are  a  dollars  per  day 
earn  in  b  days  ?  in  c  days  ?   in  a  days  ?   in  a  days  ? 

10.  If"  a  man  earns  a  dollars  per  month  and  his  expenses  are 
b  dollars  per  month,  how  much  will  he  save  in  a  year  ? 

11.  How  far  can  a  wheelman  ride  in  a  hours  at  the  rate  of 
b  miles  an  hour  ?  How  far  will  he  have  ridden  after  a  hours,  if 
he  stops  c  hours  of  the  time  to  rest  ? 

12.  A  has  X  dollars  and  B  y  dollars.  If  A  gives  B  m  dollars, 
how  much  will  each  then  have  ? 

13.  The  number  25  may  be  written  20  +  5.  Write  the  number 
whose  tirst  digit  is  x  and  second  y. 


70  MULTIPLICATION 

14.  A  book  contained  x  pages.  If  they  averaged  y  lines  to  a 
page  and  z  letters  to  a  line,  how  many  letters  were  there  in  the 
book? 

15.  Hov/  many  square  rods  are  there  in  a  square  field  one  of 
whose  sides  is  2  &  rods  long  ?   x  rods  long  ? 

16.  What  is  the  number  of  square  rods  in  a  rectangular  field 
whose  length  is  30  rods  and  width  20  rods  ?  What  will  be  the 
area  in  square  rods,  if  the  length  is  a  rods  and  the  width  6  rods  ? 

17.  A  fence  is  built  across  a  rectangular  field  so  as  to  make 
the  part  on  one  side  of  the  fence  a  square.  If  the  'field  is  a  rods 
long  and  b  rods  wide,  what  is  the  area  of  each  part  ? 

18.  A  man  who  had  s  dollars  bought  h  bales  of  hay  at  n  cents 
a  bale  and  a  bushels  of  oats  at  m  cents  a  bushel.  How  many 
cents  had  he  left  ? 

19.  A  speculator  bought  a  car  loads  of  wheat  at  m  dollars  a 
car,  and  sold  h  car  loads  of  it  at  n  dollars  a  car.  How  much  did 
he  gain  by  the  transaction,  if  he  sold  the  rest  of  the  wheat  for 
c  dollars  a  car  ? 

20.  A  sold  a  farm  which  had  cost  him  n  dollars  an  acre  to  two 
men,  a  acres  to  one  and  h  acres  to  the  other,  at  the  uniform  price 
of  m  dollars  an  acre.     How  much  did  he  gain  ? 

21.  In  a  library  there  were  p  +  q  volumes  that  averaged  p  +  q 
pages  per  volume,  p  -\-  q  words  per  page,  and  7  letters  per  word.' 
How  many  letters  were  there  in  all  these  books  ? 

22.  A  wheelman  who  had  a  journey  of  x  miles  to  make  rode 
the  first  a  hours  at  the  rate  of  h  miles  an  hour,  when  he  was 
obliged  to  stop  c  hours  for  repairs.  After  that  he  rode  2  miles 
an  hour  faster,  so  that  he  made  the  whole  journey  in  10  hours. 
What  was  the  length  of  the  journey  ? 

23.  A  wheelman  rode  a  hours  at  the  rate  of  m  miles  an  hour, 
then  decreased  his  speed  5  miles  an  hour  for  3  hours,  and  finished 
his  journey  in  6  hours  more,  increasing  his  first  rate  2  miles  an 
hour.     How  far  did  he  ride  ? 

If  a  =  4,  6  =  2,  and  m  =  10,  how  many  miles  did  he  ride,  and 
in  what  time  did  he  accomplish  the  journey  ? 


DIVISION 


101.   1.  Since  +  2  times  +  10  is  +  20,  if  +  20  is  divided  by 
+  10,  what  is  the  sign  of  the  quotient  ? 

2.  What  is  the  sign  of  the  quotient  when  a  positive  number 
is  divided  by  a  positive  number  ? 

3.  Since  +  2  times  -  10  is  -  20,  if  -  20  is  divided  by  -  10, 
what  is  the  sign  of  the  quotient  ?  if  —  40  is  divided  by  —  5  ? 

4.  What  is  the  sign  of  the  quotient  when  a  negative  number  is 
divided  by  a  negative  number  ? 

5.  What  is  the  sign  of  the  quotient  when  the  dividend  and 
divisor  have  like  signs  ? 

6.  Since  +  2  times  —  10  is  —  20,  if  —  20  is  divided  by  +  2, 
what  is  the  sign  of  the  quotient  ?  if  —  20  is  divided  by  +  5  ? 

7.  Since  -  2  times  -  10  is  +  20,  if  +  20  is  divided  by  -  2, 
what  is  the  sign  of  the  quotient  ?  if  +  20  is  divided  by  —  4  ? 

8.  What  is  the  sign  of  the  quotient  when  the  dividend  and 
divisor  have  unlike  signs  ? 

9.  How  many  times  is  2  a  contained  in  6  a  ?  in  10  a? 
How  is  the  coefficient  of  the  quotient  obtained  ? 

10.  Since  a^  x  a^  =  a^,  if  a^  is  divided  by  a",  what  is  the  quo- 
tient ?     What  is  the  quotient,  if  a*  is  divided  by  a^  ? 

What  is  the  quotient,  if  b''  is  divided  by  &^  ?  by  6*  ? 

How  is  the  exponent  of  a  number  in  the  quotient  obtained  ? 

11.  What  is  the  exponent  of  a  in  the  quotient  of  a*  -i-  a*?  of 
a'  -j-  a^  ?    How  many  times  is  a*  contained  in  a*  ?  a^  in  a^  ? 

12.  What  is  the  value  of  any  expression  whose  exponent  is  0  ? 

71 


72  DIVISION 

102.  In  multiplication  two  numbers  are  given  and  their  product 
is  to  be  found.  The  inverse  process,  finding  one  of  two  numbers 
when  their  product  and  the  other  number  are  given,  is  called 

Division. 

10  -^  2  =    5  and  D  -^  d  =  q 

are  inverses  of  5  x  2  =  10  and  q  x  d  =  D. 

The  dividend  corresponds  to  the  product,  the  divisor  to  the  mul- 
tiplier, and  the  quotient  to  the  multiplicand.  Hence,  the  quotient 
may  be  defined  as  that  number  ivhich  multiplied  by  the  divisor 
produces  the  dividend. 

The  quotient  of   a  divided  by  b,  indicated   by  (a  -^  b),  or  -, 

is  defined  for  all  values  of  a  and  b  by  the  relation 

-  X  b  =  a. 
b 

103.  Principles.  —  1.  Law  of  Signs.  —  The  sign  of  the  quotient 
is  +  when  the  dividend  and  divisor  have  like  signs,  and  —  when 
they  have  unlike  signs. 

2.  Law  of  Coefficients.  —  The  coefficient  of  the  quotient  is  equal  to 
the  coefficient  of  the  dividend  divided  by  that  of  the  divisor. 

3.  Law  of  Exponents.  —  TJie  exponent  of  a  letter  in  the  quotient 
is  equal  to  its  exponent  in  the  dividend  diminished  by  its  exponent  in 
the  divisor. 

An  expression  whose  exponent  is  0  is  equal  to  1. 

The  Law  of  Signs  may  be  established  as  follows : 

Since  +  a  x  +  b  =  +  ab,  +  ab  -^  +  b  =  +  a. 

Since  +  a  X  —  b  =  —  ab,  —  ab  -. —  b  =  -{■  a. 

Since  —  a  x  +  ft  =  —  aft,  —  aft  -f-  +  6  =  —  a. 

Since  — ax  —  6=+ aft,  +a6^ —  ft=—  a. 

The  Law  of  Exponents  or  the  Index  Law  for  Division  may 
be  established  as  follows,  m  and  n  being  positive  integers  and  m 
being  greater  than  n : 

By  §  24,        a"  =  a  X  a  X  a  •••  to  m  factors, 
a"  =  a  X  a  X  a  •••  to  ?i  factors  ; 
.-.  a"»  -4-  a"  =  (a  X  a  X  a  •••  to  m  factors)  -h-  (a  x  a  x  a  •••  to  n  factors) 
=  a  X  a  X  a  •  •  •  to  (m  —  n)  factors. 
Hence,  a""  -i-  a^  =  o*"-". 


DIVISION  73 

104.  Commutative,  Associative,  and  Distributive  Laws  for  Division. 

1.  The  Commutative  Law  may  be  established  as  follows: 
Since,  by  the  definition  of  division,  §  102,  a  =  a  -^  c  x  c, 

axb-7-c  =  a-^cxcxb-i-c 
§82,  =  a  -i-  c  X  b  X  c-i-c 

=  a  ^  c  X  b.  (1) 

Also,  §  102,  o-7-6-T-c  =  a-f-cxc-^6^c 

by  notation,  §  29,  =(^a-^cxc-^b)-i-c 

by  (1),  =(a -T-c-i- ft  X  c)-f-c 

by  notation,  =a-i-c-^bxc-i-c 

=  a^c^b.  (2) 

The  Commutative  Law  for  division  is  expressed  by  (2). 

(2)  may  be  written  -  -;-  c  =  -  -h  6.       (1)  may  be  written  —  =  -  x  6. 
be  c      c 

It  follows  from  (1)  and  (2)  that  in  a  succession  of  multiplications  and 

divisions  the  multipliers  and  divisors  may  be  arranged  in  any  order. 

2.  The  Associative  Law  may  be  established  as  follows: 
By  the  Commutative  Law  just  proved, 

axb  -^  c  =  b  ^  ex  a 

by  notation,  §  29,  ={b  ^  c)x  a 

by  the  Commutative  Law,  =  a  x  (6  -=-  c).  (3) 

Also  a-4-6-^c  =  6xc-4-(6xc)xa-i-6-f-c 

by  the  Commutative  Law,  —b^bxc^cxa-^ipxc) 

=  a^{b  xc).  (4) 

The  Associative  Law  for  division  is  expressed  by  (4). 

(4)  may  be  written  -  -r-  c  =  —  •      (3)  may  be  written  —  =  a  x  — 
b  be  c  c 

It  follows  from  (3)  and  (4)  that  in  a  succession  of  multiplications  and 

divisions  the  multipliers  and  divisors  may  be  grouped  in  any  manner,  each 

element  keeping  its  own  sign,  x  or  -i-.,  if  the  sign  x  precedes  the  sign  of 

grouping,  but  changing  it,  if  the  sign  -f-  precedes  the  sign  of  grouping. 

3.  The  Distributive  Law  may  be  established  as  follows : 


§85,                       {a^m  +  b-i-m)xm  =  a-^mxm  +  b-. 

-  m  xm 

=  n  +  6. 
Dividing  both  members  by  m, 

a-i-m  +  b-^m=:(a  +  b)-i-m; 

atis,                                                 «  +  &^.«^6. 

m        mm 

74  DIVISION 

105.    The  Reciprocal  of  a  number  is  1  divided  by  the  number. 
The  reciprocal  of  5  is  -;  of  6,  -  ;  of  (a  +  6), 


6  h  a  +  b 

106.  a-i-  b  =  a  xl  -i-b 
by  the  Associative  Law,  =  a  x  (1  -^  6). 

Hence,  dividing  by  a  number  is  equivalent  to  multiplying  by  the 
reciprocal  of  the  number. 

107.  To  divide  a  monomial  by  a  monomial. 

Examples 

1.   Divide  -ISa'b^  by  6ab^ 

Explanation.  —  Since  the  dividend  and  divisor  have 
PKOCESS  unlike  signs,  the  sign  of  the  quotient  is  —  (Prin.  1). 

6  ab^)  -  18  a^b^  Then,  -  18  divided  by  6  is  -  3  (Prin.  2);  a^  divided 

3a<5        by  a  is  a*  (Prin.  3);  and  b^  divided  by  b^  is  b  (Prin.  3). 

Therefore,  the  quotient  is  —  3  a*b. 

Rule.  —  Divide  the  numerical  coefficient  of  the  dividend  by  the 
numerical  coefficient  of  the  divisor,  and  to  the  result  annex  the  letters, 
each  with  an  exponent  eqital  to  its  exponent  in  the  dividend  minus 
its  exponent  in  the  divisor. 

Write  the  sign  +  before  the  quotient  when  the  dividend  and  divisor 
have  like  signs,  and  the  sign  —  when  they  have  unlike  signs. 


2. 

3. 

4. 

5. 

6. 

Divide 

12  a; 

-120^^/ 

35  xy^ 

-14x3^2 

-  26  a'b'c' 

By 

4x 

4aV 

-7xy 

-2x^y 

13  ab-c 

Find  the  quotient  of 

7.  2S a*b-c^ -7 abc.  10.  -  U a^f:i* -i- 7 a^fz\ 

8.  —  16  x^y^^  -=-  4  xyz.  11.  —  27  m'n^p^  -?-  9  m^n^p^. 

9.  -  36  a^mV --- 9  mV.  12.  -  3^  q^r'p  ^  -  13  qp. 

108.    To  divide  a  polynomial  by  a  monomial. 

Examples 
1.    Divide  5  x*y*  —  IQ  s^if  +  h  x^rf  by  bx^j^. 

PKOCESS.  §  104,  3,     5  a?f)5  xY  -  10  s^y"  +  5  x^f 

x^y  —    2xy^  -f       t/* 


DIVISION  75 

Rule.  —  Divide  each  term  of  the  dividend  by  the  divisor,  and  Jind 
the  algebraic  sum  of  the  jjartial  quotients. 

Find  the  quotient  of 

4,a^W-\2a%-  +  \%a*b  ^     Amhi-Smhi^ -^  4:mn^ 

4  a^b  4  mn 

24  a%-  +  32  a'W  -  40  a*b*  5  x*y  -  10  xY  +  20  x'f 

8a'b^  '  '  5xy 

-  35  a^y^z^  +  45  xyz-  ^     -a-&-c-(?-e 

5  a^y'z  —  1 

—  39  oc^yV'  +  65  x^y^z'  —a  —  a-b  —  a^c  —  a*d  —  a^e 

—  I'Sx^yh^  '  —a 

25  7-«/  _  125  ?V  -  75  r^s'" 


10. 


11. 


5  ?-V 


3  ckl- 

12.  (34  a^xV  -  51  a*xy  -  68  a^'x'y^)  -i- 17  a'x^. 

13.  (8  a'b^  -  28  a«6^  -  16  a'b'  +  4  a^6«)  -=-  4  a*ly^. 

14.  [a  (6  -  c)3  -  &  (&  -  c)-  +  c(&  -  c)]  --  (6  -  c). 

15.  [(a;  -  ^j)  -  3  (x  -  yf  +  i  x(x  -  yf]  ^(x-  y). 

16.  {X'  -  2  X''+'^  -  5  a;''+2  _  a.a+3  _,_  3  ^+4^,  ^  ^a 

17.  (^y«+l  _  2  ?/»+2  -  7/»+3  _  3  7f+*  +  ?/"+')  --  ?/"  +  '. 

18.  (a;"  —  x"-^  +  x"-^  —  ic"-^  +  x""*  —  a;""^)  -=-  a^. 

109.    To  divide  a  polynomial  by  a  polynomial. 

Examples 
1.    Divide  3  a^  +  35  +  22  a;  by  x  +  5. 


PROCESS 

3ar*+22x  +  35 
.     3a^  +  15a; 

7a;  +  35 
7  a; +  35 

a;  +  5 

TEST 

+  60  -1-  +  6 

a;(x  +  5)     . 

3«  +  7 

=  +  10 

7(a;  +  5) 

76  DIVISION 

Explanation. — For  convenience,  the  divisor  is  written  at  the  right  of 
the  dividend,  and  both  are  arranged  according  to  the  descending  powers  of  x. 

Since  the  dividend  is  the  product  of  the  quotient  and  divisor,  it  is  the 
algebraic  sum  of  all  the  products  formed  by  multiplying  each  term  of  the 
quotient  by  each  term  of  the  divisor.  Therefore,  the  term  of  highest  degree 
in  the  dividend  is  the  product  of  the  terms  of  highest  degree  in  the  quotient 
and  divisor.  Hence,  if  Sx'^,  the  first  term  of  the  dividend  as  arranged, 
is  divided  by  x,  the  first  term  of  the  divisor,  the  result,  3  x,  is  the  term  of 
highest  degree,  or  the  first  term,  of  the  quotient. 

Subtracting  3x  multiplied  by  (x  +  5),  or  3x  times  (x  +  5)  from  the  divi- 
dend, the  remainder  is  7  x  +  35. 

Since  the  dividend  is  the  algebraic  sum  of  the  products  of  each  term  of  the 
quotient  multiplied  by  the  divisor  and  since  the  product  of  the  first  term  of  the 
quotient  multiplied  by  the  divisor  has  been  canceled  from  the  dividend, 
the  remainder,  or  new  dividend,  is  the  product  of  the  other  part  of  the  quotient, 
multiplied  by  the  divisor. 

Proceeding,  then,  as  before,  7  x  -^  x  =  7,  the  next  term  of  the  quotient. 
7  multiplied  by  (x  +  5),  or  (x  +  5)  multiplied  by  7  equals  7x  +  35.  Sub- 
tracting, there  is  no  remainder.  Hence,  all  of  the  terms  of  the  quotient  have 
been  obtained,  and  the  quotient  is  3  x  +  7. 

Test.  — Let  x  =  1. 

Dividend  =  3  x2  +  22  x  +  35  =  3  +  22  -f  35  =  +  60. 

Divisor  =x-|-5  =1  +  5  =+6. 

Quotient  should  be  equal  to  +  10. 

Quotient  =3x  +  7  =3  +  7  =+10. 

Similarly,  the  result  may  be  tested  by  substituting  any  other  value  for  x. 
When  the  value  substituted  for  x  gives  the  result  0  -^  0  or  0  for  a  divisor, 
some  other  value  should  be  tried. 

Rule.  —  Arrange  both  dividend  and  divisor  according  to  the 
ascending  or  the  descending  poioers  of  a  common  letter. 

Divide  the  first  term  of  the  dividend  by  the  first  term  of  the  divisor, 
and  write  the  resxdt  for  the  first  term  of  the  quotient. 

Multiply  the  whole  divisor  by  this  term  of  the  quotient,  and  sub- 
tract the  product  from  the  divideird.  The  remainder  will  be  a  new 
dividend. 

Divide  the  new  dividend  as  before,  and  continue  to  divide  in  this 
way  until  the  first  term  of  the  divisor  is  not  contained  in  the  first 
term  of  the  new  dividend. 

If  there  is  a  remainder  after  the  last  division,  write  it  over  the 
divisor  in  the  form  of  a  fraction,  and  add  the  fraction  to  the  part 
of  the  quotient  previously  obtained. 


DIVISION 


11 


2.   Divide  2a* -ba?h  +  %o?h'' -^aW  +  h*'  by  a?-ah  +  h^ 


PROCESS 


4a63+5* 


2  a^  -  3  a6  +  6^ 


2  a^  -  5  a^ft  +  6  a?h^ 
2a*-2a?h  +  2  a?b^ 

-  3  a^d  +  4  d'b''  -  4  aft^ 

-  3  a^6  +  3  a?b''  -  3  a6^ 

a26^  -     aW  +  6* 
^252  -     db^  +  6* 

3.   Divide  a*  +  9a2  4-81  by  a2_3a-f9. 


a6  +  62 


TEST 

=  0 


a*  +  9  a^  +  81 
a4_3a3_^9a2 


PROCESS 

g^  -  3  g  +  9 


3  a'  +  81 

3  g'^  -  9  g'  4-  27  g 


g'  +  3  g  +  9 


TEST 

91-5-7 
=  13 


9  g2  -  27  g  -+-  81 
9a' -27  a +  81 

Divide,  and  test  the  results : 

4.  x^  —  x  —  20  by  ic  —  5. 

5.  a^ -I- 7  a; -h  12  by  .^-^3. 

6.  m'^  —  3  m  —  18  by  m  —  6. 

7.  ;4_6Z2_16  by  V-\-2. 

8.  cd.-0?  +  2(?  hy  c-\-d. 

9.  cc2  _  11  a;  +  10  by  a;  -  10. 

10.  x'  +  lbx  +  5A:  \>Y  x  +  6. 

11.  r^"  -f- 11  r"  +  30  by  r"  +  6. 

12.  gW  —  4  gm^  +  3  m^  by  gm  —  1. 

13.  6g2  +  13a&  +  662  by  3g  +  26. 

14.  g^  +  16  +  4  g2  by  2  g  +  g"  +  4. 

15.  g*  +  g^  +  g*  +  g2  +  3  g  —  1  by  g  +  1. 

16.  20aj22/-25a^-182/3  +  27ay^  by  62/-5». 


78  DIVISION 

17.  ay?  —  aV  —  hx^+h^  by  ax  ~  b. 

18.  a* -41  a -120  by  a^  +  Aa  +  S. 

19.  x^-61x-60  by  ar'  -  2a;  -  3. 

20.  25ar'-a;3_g^_2a..2  by  5ar'-4a;. 

21.  4^*-9?/2  +  6?/-l  by  22/2  +  3y-l. 

22.  a*  -Aa^x  +  6 a-x-  —  4 aa;^  +  a;*  by  a^  —  2 aa;  +  a^. 

23.    a''  -  1 

a^  +  a* 


—  a' 


a+  1 

-a  +  l+   ~^, 
a+1 

a*-a^  +  a^- 

a^-l 

a^  +  a^ 


a-1 
a  +  1 


24.    a:^  +    x  -  25 
a:3-3a;2 


3x2+    aj 
3ar'-9a; 

10  a;- 
lOx- 

-25 
-30 

-2 
a;-3 


ar'  +  3a;  +  10  + 


25.    a;<-3«3+    a^  +  2a;-l 

aj*  —    a^  —  2  a^ 


-2a^  +  3ar'  +  2a; 
-2a;3  +  2a^  +  4a; 

ar'-2a;-l 
a?-    x-2 


x"- 

-    x-2 

-x  +  1 
-x~2 

^- 

-2a;  +  l+-f 

XT 

x  +  l 


DIVISION 


79 


Divide : 

26.  trv'  -\-n^  by  m-\-n. 

27.  a^  +  32  by  x  +  2. 

28.  ^  +  y^  by  x^  +  y^. 

29.  a*'  +  5a*-a='  +  2a  +  3  by  a-1. 

30.  2  n'  -  4  w*  -  3  n=^  +  7  n''  -  3  n  -I-  2  by  n  -  2. 

31.  a?  + 1^ -{- ^  —  S  xyz  hy  X -\- y -\- z. 

32.  m^  +  n^  +  ar*  +  3  m^n  +  3  mn^  by  m  +  n  +  aj. 

33.  a^  -  6  a^  +  12  a  -  8  -  6"  by  a  -  2  -  6. 

34.  2/^  +  32/^  +  5y3  +  3y2  +  32/  +  5  by  y  +  1. 

35.  2a^,-x^  +  2a;*-a^  +  ar*  +  5  by  ic  +  1. 

36.  x'  +  2^-2x*  +  2ii?-l  by  a; 4-1. 

37.  a''  —  2  a^c  +  4  ac?  —  aa^  —  4  c^ic  +  2  ca^  by  a  —  a;. 

38.  a^  -  6^  +  c^  +  3  a6c  by  a^  -|-  6^  _|.  (.2  _^  ^5  _  ^^  +  he. 

39.  a;"  4-  ?/"  by  x-{-y  to  five  terms  of  the  quotient. 

40.  a^  —  6  a;  +  5  by  a^  —  2  a;  +  1,  using  detached  coefficients. 


1+0+0+0+0-6+5 
1-2  +  1 


PROCESS 

1-2  +  1 


2- 

-1+0 

2- 

-4  +  2 

3-2  +  0 

3-6-r.5 

4- 

-3- 

-6 

4- 

-8  +  4 

5- 

-10  +  6 

5- 

-10  +  5 

1+2+3+4+5 


a:(*  +  2a!*  +  3ar^  +  4x  +  5 


80  DIVISION 

Divide,  using  detached  eoefl&cients  when  convenient : 

41.  a;^  +  8a;  +  7  by  a^  +  2ic  +  l. 

42.  a*' -f  38  a +12  by  a +  2. 

43.  m''  — 19  m  — 6  by  m-\-2. 

44.  m^  -  32  m^  -  4  m  +  8  by  m  -  2. 

45.  a*'  +  27a2-9a-10  by  a^-3a  +  6. 

46.  21a;*  -29a;3-8a,'2+6a;  +  4  by  3ic-2. 

47.  2a;*-lla^  +  16ic2-12a;  +  9  by  2a;-3. 

48.  30.T*-62a^  +  60a^-36a;  +  8  by  5a;-2. 

49.  21x*-SSo(?  +  ^Qx'-lVdx  +  b5\iy  9x-5. 

50.  x^  -2x'-a?~10x-3Q  hy  x-2. 

51.  «*  — 4a;3+ 5a^  — 4a;  +  l  by  a;^  — ic  +  l. 

52.  x^-^-lOx'  +  lx  +  l^hj  x^-2x-3. 

53.  2a^  +  7a^- 27 a^-8a;  + 16  by  a^  +  5a;- 4. 

54.  28a;*  +  6a^  +  6a;2-6a;-2  by  4ic2  4-2a;  +  2. 

55.  7a^-6iK*  +  2a;2_a,_2  by  6ar=  +  5a;  +  2. 

56.  25x'-20a?  +  Zx*  +  Ux-&hj  Sx'-%x  +  2. 

57.  3a;*  +  7a^  +  6ar^  +  3a;-l  by  ar'  +  x  +  l. 

58.  6a^-23a:3  +  30a.-2-18a;  +  4  by  2i»2_5^_^2. 

59.  24a;*  +  32a^-16a^-25a;-4  by  6ar^-a;-4. 

60.  ar*-2a;*  +  T^a^  +  |a^  +  yV^'+f  by  a;-f. 

61.  ar^  -  f  a*  +  If  a^  -  l^or' +  f  a;  -  ^  by  a;  -  |. 

62.  a*-iar>  +  |aJ«-^a^  +  e^-H^  +  A  by  ^-^x+h 

63.  |a^  +  ^^  +  z3-^a;2/2  by  |a;  +  ^2/-|-a. 

64. to  five  terms. 

1  +  a; 

65.  to  five  terms. 

1  ~  X 

66.  a;'"-^  +  2/^"+^  by  x"~^  +  y"'*'\ 


DIVISION 


81 


SPECIAL   CASES   IN   DIVISION 
110.   By  actual  division, 


x-y 
a?  —  1^ 

x-y 

X*  —  y* 
x-y 

I  x-y 


=  x-\-y, 

=  x^  +  xy+y'. 

=  Qc^  +  x'y  +  xy^  +  y*. 

=  X*  -\-  x^y  +  x^y^  +  xy^  +  y*. 


From  the  above  we  infer  that  the  difference  of  the  same  powers 
of  two  numbers  is  divisible  by  the  difference  of  the  numbers. 


x-\-y 
^  —  'i^ 

x  +  y 

x^-y 
x  +  y 


=  x-y. 

=  x^  —  xy  -{-y^,    Rem.,  —  2  2/*. 

=  3?  —  x'y  +  xy^  —  ?/*. 

=  x* —  !3?y +  ci?y^ —  xy^  ^y^,    Rem.,  —2?/*. 


From  the  above  we  infer  that  the  difference  of  the  same  powers 
of  two  numbers  is  divisible  by  the  sum  of  the  numbers  only  when 
the  powers  are  even. 

{x'  +  y'' 

^x  +  y,    Rem.,  22/^ 

=  2i?-\-xy  +  y^,     Rem.,  2  f. 

=  21?  +  x^y  +  xy^  +  'f,     Rem.,  2  y*. 

^_     =x*  +  x'y  +  a?y'-  +  xy^  +  y*,     Rem.,  2?/*. 

From  the  above  we  infer  that  the  sum  of  the  same  powers  of 
two  numbers  is  not  divisible  by  the  difference  of  the  numbers. 

ALG.  6 


3. 


x  —  y 
^  +  f 

x-y 

x^  +  y' 

x-y 

82 


DIVISION 


01?  +  y- 
^  -|— ,-r  =  ^  —  y>    Rem.,  2  y\ 


a^  +  y^ 
a^  +  ?/^ 


=  a^  -  .Ti/  +  y''. 

=  u?  —  x^y  -\-  xif  —  y^,     Rem.,  2  ?/^ 

=  a;''  —  ar'y  4-  a;y  —  xy^  +  r/*. 


Observe  that  the  sum  of  the  same  powers  of  two  numbers  is 
divisible  by  the  sum  of  the  numbers  only  when  the  powers  are  odd. 

111.  Hence,  when  n  is  a  positive  integer. 
Principles.  —  1.   x^  —  y^  is  always  divisible  by  x  —  y. 

2.  x"  —  y"  is  divisible  by  x  -\-  y  only  when  n  is  even. 

3.  xf  -f  2/"  is  never  divisible  by  x  —  y. 

4.  ic"  +  y"  is  divisible  by  x  +  y  only  when  n  is  odd. 

112.  From  §  110,  the  following  law  of  signs  may  be  readily 
inferred : 

When  x  —  y  is  the  divisor,  the  signs  in  the  quotient  are  x>lus. 
When  x  +  y  is  the  divisor,  the  signs  in  the  quotient  are  alternately 
plus  and  minus. 

113.  The  following  law  of  exponents  may  also  be  inferred : 
When  a;"  ±  ?/"  is  divided  by  x  ±y,  the  quotient  is  homogeneous, 

the  exponent  of  x  decreasing  and  that  of  y  increasing  by  1  in  each 
successive  term. 

114.  Proofs  of  preceding  principles. 

Fbinciple  1 


1st  Rem., 
2(1  Kern., 


a;"- 

-r 

X"- 

-oc"- 

'y 

x»- 

'y- 

yn 

a!»- 

'y- 

Xn- 

-2y2 

x-y 


+  x»--y  + 


ji;n-2^2  _  yn 


nth  Rem., 


Q^n-nyn  _  yn 

xOyn  -  yn  -yn  ^  yn  =  Q 


DIVISION  83 

By  dividing  until  several  remainders  are  obtained,  it  is  found  that  the 
first  term  of  the  first  remainder  is  a;»-'y  ;  of  the  second,  x''-^  J  of  the  third, 
a;n-3y8 .  of  the  fourth,  x"-*j/* ;  and  consequently  of  the  nth,  x"-"?/".  But 
x«-"  =  x",  which,  §  103,  equals  1.  Therefore,  the  first  term  of  the  «th  re- 
mainder reduces  to  y". 

Since  the  second  term  of  the  nth  remainder  is  —  j/",  the  entire  nth  re- 
mainder is  ?/»  —  2/»,  or  0  ;  that  is,  there  is  no  remainder,  and  the  division  is 
exact. 

Therefore,  x"  —  y"  is  divisible  hy  x  —  y  when  x  and  y  represent  any  two 
numbers  and  n  is  any  positive  integer. 


Principle  2 


x"  —  y" 
X"  +  x"-ij/ 

1st  Rem.,  —  x^-iy  —  j/» 

—  x"-'j/  —  x^-h/"^ 


2d  Rem.,  x"-2y2  _  ^n 

x"-2j/2  +  x^-Sy* 

3d  Rem.,  -  x^-^y^  -  y« 

—  x"~^^  —  x"~*y* 

4th  Rem.,  x"-V  -  y" 

Suggestion.  —  Since  the  second  term  of  each  remainder  is  negative,  no 
remainder  can  reduce  to  0  unless  its  first  term  is  positive.  Show  for  what 
values  of  n  such  remainders  reduce  to  0  when  n  is  a  positive  integer. 

Prove  Principle  3. 

Prove  Principle  4. 

Examples 

Write  by  inspection  the  quotient  of 
1.    1.^^.  6.    ^'-<  11. 


12. 


a? 

-b^ 

a 

-b 

mP 

-f-n" 

m 

+  n 

^ 

-./ 

X 

+  y 

r' 

-s' 

r 

—  s 

a> 

+  b' 

m  - 

-n 

n'- 

1 

n  — 

1 

&- 

d? 

c  — 

d 

1  + 

a» 

1  -\-a 

0*- 

1 

13. 


14. 


a;2-9 

a;  +  3 

a^-S 

a-2 

ar*-32 

a;-2 

c3  +  27 

c  +  3 

a'  4- 128 

10.    ^^^ ^-  15. 

a  +  6  a;.-j-l  a  +  2 


84  DIVISION 

16.  Find  five  exact  binomial  divisors  of  a^  —  «*. 

Solution 

««  —  a*  is  divisible  by  a  —  a;  (Prin.  1). 

0(6  _  a«  is  divisible  by  a  +  x  (Prin,  2). 

Since  a^  —  x^  =  (^a^y  _  (x^)^,  a^  —  ofi  may  be  regarded  as  the  difference  of 
tv70  odd  powers,  and  is,  therefore,  divisible  by  a?  —  x^  (Prin.  1). 

Since  a^  —  ifi  —  {a^Y  —  (x3)2,  a^  —  x®  may  be  regarded  as  the  difference  of 
two  squares,  and  is,  therefore,  divisible  by  a?  —  x^  (Prin.  1). 

Since  a^  —  x^  =(a^y  —  (x^)^,  a^  —  x^  may  be  regarded  as  the  difference 
of  two  squares,  and  is,  therefore,  divisible  by  a^  +  x^  (Prin.  2). 

Therefore,  the  exact  binomial  divisors  of  a^  —  x^  are  a  —  x,  a  +  x,  a^  —  x^, 
a^  —  x^,  and  a^  +  x^. 

17.  Find  an  exact  binomial  divisor  of  a®  +  a^. 

Solution 

Since  a^  +  x^  =(0^)^  +(x^y,  a^  +  ofi  may  be  regarded  as  the  sum  of  the 
cubes  of  a^  and  x^,  and  is,  therefore,  divisible  by  o^  +  x^  (Prin.  4). 

Find  exact  binomial  divisors  : 

18.  a^  —  ml  24.  ic^  +  a\  30.  a*  -  6*,  four. 

19.  a^-m^  25.  a}^ -\- b^'^.  31.  a«  -  1,  five. 

20.  b^  +  x^.  26.  a}"  +  6*.  32.  a^  —  6^  six. 

21.  «^- a'.  27.  ai2_,_5i2^  33.  a^o  _  ?,io^  five. 

22.  c^-hn\  28.  a^  -  27.  34.  a^®  -  6^^  eight. 

23.  a«4-6^  29.  a*'-27.  35.  a^^  _  512^  nine. 

Equations  and  Problems 
115.    1.   Find  the  value  of  a;  in  the  equation  603  —  6^  =  ca  —  c^. 

Solution 

bx  —  b^  =  cx  —  c'. 
Transposing,  bx-cx  =  b^  -  c^. 

Collecting  coefficients  of  x,      (b  —  c)x  =  ft'-^  —  c^. 

52  _  g2 

•  Dividing  by  6  -  c,  x  =  — =  0  +  c. 


DIVISION  85 

2.   Find  the  value  of  x  in  the  equation  x  —  o?  =  2  —ax. 

Solution 
X  —  a^  =  2  —  ax. 
Transposing,  ax  +  a;  =  a^  +  2, 

Collecting  coefficients  of  x,      (a  +  l)x  =  o^  +  2. 

Dividing  by  a  +  1,  x  =  ^!-±-?  =  a2  _  ^  +  j  ^ . 


a+ 1  a+ 1 

Solve  the  following  equations : 

3.  1  a  —  10  =  a?  —  ax-\-5x.       6.    co;  —  c^  —  d^ -f- da;  =  0. 

4.  a;  —  1  —  c  =  ca;  —  c^  —  c*.        7.    a^  --  ax  —  2  ab  -\-  hx  +  h^  =  0. 

5.  2m^  — ma;-f wa;  — 2w^  =  0.     8.    2%"^ -\-bn-{-x  =  n^ —  nx  —  2. 

9.    ri^a;  —  3  7n?n^  -\-  nx  -\-  Z  w?  +  x  =  0. 

10.  a^a;  -  a'' +  ^a^  +  Sa;- 5a +  10  =  0. 

11.  Zah-  a^  -2hx  =  2b'^  -  ax. 

12.  9  a^  4-  4  ma;  =  -  (3  aa;  -  16  m^). 

13.  c?/-c^-2c3-2c2  =  2c-y  +  l. 

14.  ay  —  2by  +  3cy  =  a  —  2b  +  3c. 

15.  z  +  Gw*  — 4n^  =  l  —  3n2!-)-2w  — nl 

16.  X  -  3b^ -192  b^c^- 4. ex  +  16 c'x  =  0. 

17.  8&3-1862_576-2&a;  +  7a;  +  77  =  0. 

Solve  the  following  problems : 

18.  A  drover,  who  had  5  times  as  many  sheep  as  oxen  and  ^  as 
many  oxen  as  horses,  sold  all  for  $  2300,  —  the  horses  at  f  35  a 
head,  the  oxen  at  f  25  a  head,  and  the  sheep  at  $  4  a  head.  What 
was  the  number  of  each  ? 

19.  A  man  paid  yearly  a  certain  amount  of  money  for  taxes 
and  twice  that  amount  for  improvements,  and  received  for  rent 
3  times  as  much  as  he  paid  out  for  improvements.  If  his  net 
gain  per  year  was  $  300,  what  were  his  taxes  per  year  ? 

20.  A  owed  B  a  certain  sum  of  money  and  C  twice  as  much. 
D  owed  A  3  times  as  much  as  A  owed  B,  and  E  owed  A  5  times 
the  sum  A  owed  B.  A  found  that  if  he  could  settle  with  them  all 
he  would  have  $  5000.     How  much  did  he  owe  B  and  C  ? 


86  DIVISION 

21.  After  taking  3  times  a  number  from  11  times  the  number 
and  adding  to  the  remainder  7  times  the  number,  the  result  was 
12  less  than  117.     What  was  the  number  ? 

22.  A  merchant  failed  in  business,  owing  A  3  times  as 
much  as  B,  C  twice  as  much  as  A,  and  D  as  much  as  A  and  B. 
If  the  entire  debt  to  A,  B,  C,  and  D  was  f  28,000,  how  much  did 
he  owe  each  ? 

23.  At  a  certain  election  there  were  three  candidates  for  the 
office  of  mayor.  A  received  half  as  many  votes  as  B  and  4  times 
as  many  as  C.  If  the  total  vote  lacked  25  votes  of  being  2300, 
how  many  votes  did  each  receive  ? 

24.  Three  boys  together  had  140  marbles.  If  the  second  boy 
had  twice  as  many  as  the  first  and  half  as  many  as  the  third,  how 
many  had  each  ? 

25.  In  a  certain  school  of  600  students  there  were  twice  as 
many  Sophomores  and  3  times  as  many  Freshmen  as  Juniors, 
and  40  more  Seniors  than  Juniors.  How  many  students  were 
there  in  each  class? 

26.  Divide  25  into  three  parts  such  that  the  first  is  one  third  of 
the  second  and  5  greater  than  the  third. 

27.  A,  B,  and  C  divided  $40  so  that  for  every  $2  A  received, 
B  and  C  each  received  $  3.     What  was  the  share  of  each  ? 

28.  Divide  $2200  among  A,  B,  and  C,  so  that  B  shall  have 
twice  as  much  as  A  and  $200  less  than  C. 

29.  Divide  $  351  among  three  persons  so  that  for  every  dime 
the  first  receives  the  second  shall  receive  25  cents  and  the  third  a 
dollar. 

30.  A  man  gave  equal  amounts  of  money  to  a  school  and  to  a 
library,  and  ^  the  same  amount  to  a  hospital.  If  to  all  he  gave 
$  28,000,  what  sum  did  he  give  each  ? 

31.  When  wheat  was  worth  85  cents  a  bushel,  oats  35  cents  a 
bushel,  and  corn  60  cents  a  bushel,  a  man  bought  a  quantity  of 
wheat,  oats,  and  corn  for  $67.  If  he  bought  twice  as  many 
bushels  of  oats  as  of  wheat,  and  also  three  times  as  many  bushels 
of  corn  as  of  wheat,  how  many  bushels  of  each  did  he  buy  ? 


REVIEW 


87 


REVIEW 
116.    Simplify; 

1.  a^  +  2  aVxy  —  3  mn  +  4  mn  —  4  a^  —  5  a^/xy  +  3  «^  +  4  a^/xy 

■  2mn. 

2.  h  01?  +  3 3?y  -\- ^  xy^  — '^  —  V^  +  6  ?/^  +  a^—  Vy  —  5a?y  —  7  a? 

■  5xy^  +  x^  +  2^x  +  a^y  —  6  2/^  4-  V?/  —  Va;  —  2  a.*^?/  +  3  ajy'^  +  y?. 

3.  (f  a  -  3  6c  +  i  c  -  7  6)  -  (f  a  +  ^  6c  +  ^  c  +  3  6). 

4.  (a-ar^  —  4  ay  +  4  6c  +  ax)  —  (6V  —  4  6y  —  aa;  +  6c). 

5.  V?  -ix"  -hx'^y  ^X^ix'f  -Vd^f  ^hxy"  -f). 

6.  |a-|a;-(fa-ia:)-(36--V-a;-|a)  +  ^a. 

7.  (a^  +  3 a^^/  +  3  ay^)  (a'  -  2  ay  +  y^- 

8.  (a^"  +  2  a;»y»  +  y*")  (a^"  —  2  a;"?/"  +  /»). 

9.  {y^^^lxy^\xf){\7?-\xy^\y% 
10.  (.2a2-.8a  +  1.6)(.la2+.4a+.8). 

Expand : 


11.  (y-3)(2/  +  4).  21. 

12.  (y  +  7)(y-8).  22. 

13.  (.v-l)(2/  +  2).  23. 

14.  (y-5)(y-9).  24. 

15.  (y+8)(?/-4).  25. 

16.  (m  —  a;)  (m  +  x).  26. 

17.  (m4-a)(m  — 6).  27. 

18.  (a;  —  m)  (a;  +  w).  28. 

19.  (x2  _,_  j.^  (^  ^  2).  29. 

20.  (x2  +  4)  (ar^  -  3).  30. 

31.  (a  -  6)  (a  +  6)  (a==  +  62). 

32.  (l-a;)(l+a;)(l  +  a^)(l  +  X*). 

33.  (1  — a;)(l +a;)(l— a;)(l  +  x). 

34.  (m  +  ^0  (w,  -f  w)  i^n  —  w)  (m  —  w) 


(a;  +  m)  (x  +  m). 

(af'  +  ic^)  (a;  +  1). 

(a;-l)(l+x). 

(a;  +  3i!/)(x-2y). 

(a»  +  6"')(a"-6'»). 

{or  +  6")  (a"  -  6"). 

(a  +  6  +  c)  (a  +  6  —  c). 

{X  -\-  y  ^  z){x  -  y  ^  z). 

(r  +  s  ~  t)  (s  —  t  —  r). 

{m  +  n  —p)(m  —  n  -^  p). 


88  REVIEW 

35.  (a*-\-a^  +  a^  +  a  +  1) (a -  1). 

36.  {a^  +  a^y  +  xy-  +  f)(x-y). 

37.  (ic*-a;*  +  a^-a^  +  a;-l)(a;  +  l). 

38.  (a^  +  2 a*  +  4.a^  +  80^  +  16a  +  32)(a  -  2). 

Square : 

39.  2x-oy.  42.  100-5.  45.  a  +  b  +  c  +  d. 

40.  x*  —  aa^.  43.  w^"  —  m".  46.  2  a  — 3  6  — 4  c. 

41.  50  —  1.  44.  7x  —  b^y  47.  a;"-i  -  ?/ -  al 

Expand : 

48.  (5a-4.v)(5a-3y).  51.  (2 a^a; -  5 ft^y) (4 a^x -  3 6^)- 

49.  (6x~4y)(3x  +  5y).  52.  (6 amn  +  5 j?) (6 amw  -  3p). 

50.  (3  a; +  a^)  (3  a;  4- 62/).  53.  (3a"+i-26»-i)(2a"+i- 3  6"-»). 

54.  (a;  +  2/)(a;-2/)(a^  +  y^(a^+2/')(aJ«  +  2/«). 

55.  (m8  +  1)  (m*  + 1)  (m^  + 1)  (m  +  1)  (m  -  1). 

56.  (16a;*  +  l)(4a:2  +  l)(2a;  +  l)(2a;-l). 
Divide : 

57.  ;r*  -  2  a^  +  2  a^  +  12  a;2  -  a;  -  8  by  a;  +  1. 

58.  a;*  -  4  a^  +  5  a.-2  -  4  a;  +  1  by  a;-  -  3  a;  +  1. 

59..  a^-45a^  +  45a;*-18a;2_,_9^_l  ^^  a^- 4ar'  +  3a;- 1. 

60.  a'-12a^-a  +  12  hy  a^-2a'  +  4:a-3. 

61.  6«-10&2_56  +  4  by  63_262  +  36-l. 

62.  m^"  -  6  w^  +  5  m  -  2  by  m*  +  2  m^  -  3  m  -  2. 

63.  a^-160aH127a3_i00a=^-20a+16  by  a3-6a2+5a-4. 

64.  610+29  6*-170  63-61  624.210  6-22  by  6^+2  6^-5  6-11. 

65.  6a^-HffaV-ffa/  +  |i/  by  a^  +  i  a'^y  -  i  a^/^  +  i  y^ 

66.  a^c—ab^+acd~ad^  -abc-\-b^-bcd+bd'-a(^-\-cb^-(^d-{-cd^ 
by  ac  —  6^  +  cd  —  d^ 


REVIEW  89 

67.  Divide  1  by  1  —  a;  to  six  terms. 

Simplify : 

68 .  4  a^  +  3  6^  -  (2  a^  -  3  6^)  -  (7  6^  -  2  a^  -  (  -  a'  -  6«). 


69.    a-  -  (b' _  c2)  -  (62  +  c^  -  a')  ^{c'-b'-a^-  (a" -  6^  -  c*). 


70.  a?  -  {2xy  -  y")  -  {a?  +  xy  -  y"^  -  x"  -2xy  -  y^  +  ^y\ 

71.  m  +  |2m-[7i  +  3|?-(4p-3n)-57i  +  2m]— 7p|. 


72.  X  —  d y  +  \2 z  —  {5 y  —  ^x  —  1  z)  —  {x  -  y)  —  z\  —  X. 

73.  1  -{l-[a^-3-(2a;2-4)4-3r'^l-]_(a;2_4)|-l. 

74 .  a  -  (2  6  +  5  a)  -  2  6  -  [3  a  -  (6  6  -  5  a)  -  a]  +  12  a. 

75.  x  —  \m—{x  +  {3m  —  2x)-^5x—(2x  +  m)']—2x  +  m\. 


76.    x-55a;-[6a;-(7a;-8a;-9a;)-10a5]+lla;|4-9a;. 


77.  3 -fc-[5- (2c- 7 -3c- 11) +  5c]- 6c -20|. 

78.  .T  +  13y+[4a;-(2y-7a;)-32/]-(102/  +  4a;)  +  ^y|- 

79.  1  _  I  _ [_  (1  _  a;)  -  1]- 1|  -  )x  -  (5  - 3  a;)  -  7  ^  aJ^ 

Collect  the  coefficients  of  x,  y,  and  z : 

80.  ax+  ay  +  az  —  bx  —  by  —  bz. 

81.  cue  —  2?/ +  C2:  + 6?/  — 12a;4- 42. 

82.  Smx  —  nx-{-by  —  y  +  Scz  —  4:Z. 

83.  p?/  —  y  —  42  +  62  —  a;  +  ?na;  —  wic  —  2. 

84.  ex  —  6?/  —  3  a2  4-  a;  —  y  —  4  2  +  2  —  y. 

85.  16ny —  16mx  + ax +  by +  cx  —  2y. 

86.  ma;  +  ny  +  az  +  2ax  —  2  my  +  2  n2. 

87.  X  —  y  —  az  -{-  S  mx  -\-aby  —  x^  -\- y^  -\- z. 

88.  a^a;  +  6^.v  —  2  ax  —  2  C2  +  c22  +  a;  +  y  +  «• 

89.  m^x  —  nhj  +  m^y  —  n^x  —  2  mnx  —  2  mwy  —  n''?  4-  z- 

90.  4  (ax  —  6i/  +  C2)  —  2 (6x  —  a?/  —  C2)  —  2  (x  —  y  +  z). 


FACTORING 


117.  The  numbers  that  multiplied  together  produce  a  given 
number  are  called  its  Factors. 

The  factors  of  12  a  are  2,  2,  3,  and  a;  or  4,  3,  and  a ;  or  2,  6,  and  a ; 
or  12  and  a  ;  or  2  and  6a;  or  4  a  and  3,  etc. 

118.  A  number  that  has  no  integral  factors  except  itself  and  1 
is  called  a  Prime  Number. 

119.  A  number  that  has  integral  factors  besides  itself  and  1  is 
called  a  Composite  Number. 

120.  A  factor  that  is  a  prime  number  is  called  a  Prime  Factor, 

121.  The  process  of  separating  a  number  into  its  factors  is 
called  Factoring. 

122.  To  factor  a  monomial. 

The  factors  of  the  numerical  coefficient  are  found  as  in  arith- 
metic, but  the  factors  of  the  literal  part  are  evident. 

Thus,  in  a^,  the  three  factors  are  as  evident  as  if  they  were  written  axay-a. 

It  is  seldom  necessary  to  resolve  a  monomial  into  its  simplest 
factors,  but  the  following  problem  often  occurs : 

Given  one  factor  of  a  monomial,  to  find  the  other. 

Rule.  —  Divide  the  monomial  by  the  given  factor. 

1.  In  each  of  the  following,  if  xy  is  one  factor,  find  the  other: 
6  or*?/,  15  x*y^,  2  a^?/^,  d^y^h'^y^,   —  mnxy,  —  xy. 

2.  In  each  of  the  following,  if  abc  is  one  factor,  find  the  other: 
a^hc,  ah\  abc?,   —  a%^<?,   —  a^bc,  —  \  abc. 

3.  Find  two  equal  positive  factors  of  o^ ;  of  9  aV ;  of  64  m*. 

4.  Find  two  equal  negative  factors  of  25  a;*;  of  16  a*;  of  9  a*. 

90 


FA  CTORING  91 

123.  To  factor  a  polynomial  whose  terms  have  a  common  factor. 

Examples 

1.  What  are  the  factors  of  3  a?xy  —  6  aa?y  +  9  axy^  ? 

PROCESS  Explanation. — By  examining  the  terms 

3  a^xv 6  ax^v  4-  9  axi/^   °^  ^^^  polynomial,  it  is  seen  that  3  axy  is  a 

=  3  axy  (a  —  2x  +  3v)        ^^^*°^  '^^  ^"^^^^  *^^™*    ^i^i'^^'S  ^y  *^i^  ^o™' 
'^         mon  factor,  the  other  factor  is  found. 

Hence,  the  factors  are  3  axy,  the  monomial  factor,  and  (a  —  2x  +  3y^,  the 

polynomial  factor,  since,  by  the  Distributive  Law  for  multiplication, 

3  axy  (a  —  2x  +  3y)  =  3  a^xy  —  6  az^y  +  9  axt/^. 

Find  the  factors  of  each  of  the  following  polynomials : 

2.  5  x^  —  5  x^.  14.  ac  —  be  —  cy  —  abc. 

3.  80-^  +  20;*.  15.  3  a^//3  -  3  ar'/ +  12  a;?/. 

4.  3x3-6a^2/-  16.  16  a^/Zc*  -  24  a'ftV  +  32  a^&V. 

5 .  4  a^  —  6  a6.  1 7 .  60  m^n^i'^ — 45  mVr' + 90  m*n^i^. 

6.  5m2-3m7i.  18.  12  a%  -  18  a6y  +  24  a^^y. 

7.  3  if^?/^  —  3  xry^.  19.  14  a^mn^— 21  a^m^n^— 49  a*mn^. 

8.  5m%-10mV.  20.  12  afy^z^  - 16  afy^z^  -  20  a^yh^ 

9.  la^^-Ga^ftl  21.  25  c^dar' +  35  c-^dV  -  55  c^d^a^. 

10.  5x^-10a^-5a;l  22.    51  xy^z' ~68  a^fz^ +  85  xYz*. 

11.  3a*-2a'b  +  aVA  23.    52  a^ftV  -  65  a362g2  _  9^  ^2^2^ 

12.  x^2  +  X-"  +  a;'"  -  x».  24.    44  aV/ 4- 66  aVy' +  88  aVy*. 

13.  3  m^ - 12  mV+ 6  m?i*.    25.    84ary~36a^?/'-f  60a^?/*-48a;/. 

124.  To  factor  a  polynomial  whose  terms  may  be  grouped  to  show 
a  common  polynomial  factor. 

Examples 
1.    Factor  ax -\-  ay  +  bx -\-  by. 

Solution 

ax  +  ay  +  bx  +  by  =  a(x  +  y)  +  b(x  +  y) 
=  ia  +  b){x  +  y). 


92 


FACTORING 


2.  Factor  ax  —  ay  —  hx  +  by. 

Solution 
ax  —  ay  —  bx  -{■  by 
=  a(x  -y)-  b{x  -  y) 
=  ia-h){x-y). 

Observe  that,  when  the  first  two  terms  are  factored,  (x  —  y)  is  found  to 
be  the  binomial  factor.  Since  {x  —  y)  is  to  be  a  factor  of  the  other  two 
terms,  the  monomial  factor  is  —  6,  not  -f  h. 

3.  Factor  ex  +  y  —  dy  -\-  cy  —  dx  +  x. 

Solution 

GK  +  y  —  dyA-cy  —  cbc-\-x 
=  ex  —  dx  +  x  -\-  cy  —  dy  +  y 
=  (c-d+l)x  +  (c-d+ l)y 
=  (c-d+l)(a;  +  2/). 


Arranging  terms, 


Factor  the  following  ; 

4.  am  —  an  +  mx  —  nx. 

5.  bc  —  bd  +  cx  —  dx. 

6.  pq  —px  —  rq+  rx. 

7.  ay  —  by  —  ab  -f-  b\ 

8.  a?  —  xy  —  bx  +  5y. 

9.  b-  —  bc  +  ab  —  ac. 

10.  ic^  +  x'y  —  aa;  —  ay. 

11.  c^  — 4c  +  ac  — 4a. 

12.  2  a;  — ?/  + 4ar'  — 2  a;?/, 

13.  1  —  m  +  n  —  mn. 

14.  22>  +  ^  +  6p^  +  3pg 

15.  ar  —  rs  —  ab  +  bs. 

28 
29 


16.  a.-^  4- a;^  +  a;  +  1. 

17.  _j^  +  2/'-3t/-3. 

18.  a;^  +  af' 4- a;^?/ +  y. 

19.  2-2n-n''  +  n\ 

20.  .^'^  —  a;  —  a  +  aa;. 

21.  S2(?-15x  +  10y-2a?y. 

22.  12a''-8a&-3a*  +  2a''6 

23.  3  m^n  —  9  mn^  +  am  —  3  an. 

24.  loab--^bh-3oab +21  be. 

25.  16  ax-  +  12  ay  -  8  6x  -  6  by. 

26.  aa."^— aaj— aa^y  +  ay+a;— 1. 

27.  a^y+a;— 3  t/^— 3  y— 41/— 4. 
aa;  —  a  —  6a;  +  6  —  ca;  +  c. 
mx—nx—x—my+ny-\-y. 


30.   a^— a— a&+6— 2ac+2c. 


FACTORING  93 

31 .  mp^  —  np^  +  TTiq  —  nq  +  m  —  n. 

32.  aa^  —  bixr  —  ax  +  bx+a  —  b. 

33.  2  ma^  —  Tix^ -{- n -\- 2  nx  —  4^  mx  —  2  m. 

34.  bx^  —  b  —  xy  —  y-\-  ya?  —  bx. 

35.  a^x  —  a^y —  ay —  y +  x  +  ax. 

36.  2-3&  +  3a6-2a-|-4a2-6a26. 

37.  m^ -\- mn  +  mn  +  n^ -{■  m -\- n. 

125.  To  factor  a  trinomial  that  is  a  perfect  square. 

1.  Since  (a  +  b)(a  +  b)  =  a^  +  2ab  +  b^,  what  are  the  factors  of 
a^  +  2ab  +  b^?    How  are  they  obtained  from  a^  +  2ab  +  b^? 

2.  Since  (a  —  6)(a  —  b)=a^  —  2ab-\- b^,  what  are  the  factors  of 
a^  —  2ab  +  b^?     How  are  they  obtained  from  a^  —  2 a6  +  6^ ? 

3.  What  determines  the  sign  that  connects  the  terms  of  each 
factor  ? 

126.  One  of  the  two  equal  factors  of  a  number  is  called  its 
Square  Root. 

127.  In  a  trinomial  that  is  a  perfect  square  one  term  is  equal 
to  twice  the  product  of  the  square  roots  of  the  other  two  terms. 

25  a;2  —  20  xy  +  4  y"  is  a  perfect  square,  for  twice  the  product  of  the  square 
root  of  25  x^  and  the  square  root  of  4  y^  is  20  xy. 

Rule.  —  Connect  the  square  roots  of  the  terms  that  are  squares 
with  the  sign  of  the  other  term,  and  indicate  that  the  result  is  to 
be  taken  twice  as  a  factor. 

From  any  expression  that  is  to  be  factored,  the  monomial 
factors  should  usually  first  be  removed. 

Thus,  2a3-4a2  +  2a  =  2  a{a^  -2a  +  l)  =  2a(a-  1)2. 

Examples 
Factor  the  following : 

1.  x^  +  2xy  +  y\  5.   x'  +  Gx-^Q. 

2.  p^-2pq^q\  6.    m''-8m.+  16. 
2.   c'  +  2cd+d?.                             1.   x^-2x  +  l. 

4.    m^-2mn-\-n\     ■  8.    a^- 16 a +  64. 


94  FA  CTORING 

9.  ar'  +  4a;  +  4.  20.  9  +  42  6^  _|.  49  fts 

10.  4-4a  +  a2.  21.  9m^-6m^  +  l. 

11.  4a-4a''  +  a'.  22.  4 a.^ - 20 rci/ +  25. 

12.  ^x'  +  Qxy  +  ^y^  23.  4 ar' + 12 a;?/z  +  9 /2«. 

13.  2m2  — 4mw  +  2nl  24.  9aW  — 6a'm  +  l. 

14.  5ar  +  30a;  +  45.  25.  2 x  +  20 a^a;  +  50 a^x 

15.  10 a^- 20 a; +  10.  26.  18  a^6  4- 60  aft^  +  50  6'. 

16.  16p2_24i)  +  9.  27.  a''af-2ax'bf  +  bY. 

17.  9a.-2-42a;  +  49.  28.  25 a'™  -  60 a'-ft"  +  36 &*•. 

1 8.  1  +  4  &  +  4  6^.  29.  a^"  -  2  x"y"z"  +  ^^"z^". 

19.  l-6a3-f9a«.  30.  81  a'b^c?  +  IS ab'c'd -\- b'c?d\ 

When  either  or  both  of  the  squares  are  polynomials,  the  expres- 
sion may  be  factored  in  a  similar  manner. 

31.  Factor  a^  +  6x(x-y)  +  9{x-7jy. 

Solution 

352  +  6  a;  (x  -  ?/)  +  9  (x  -  y)2 
=  [x  +  3(x-y)][x  +  3(x-y)] 
=  (x  +  3x-3j/)(x  +  3x-3y) 
=  (4x-3y)(4x-32/). 

32.  Factor  (a  -  6)^  +  2  (a  -  b)(b  -c)  +  {b-  cf. 

Solution 

(a  -  6)2  +  2  (a  -  6)(6  -  c)  +  (&  -  c)2 
=  [(«-6)  +  (b-c)][(a-6)  +  (6-c)] 
=  (a-6  +  6-c)(a-6  +  6-c) 
—  (a  —  c)(a  —  c). 


Factor : 


33.  a^  +  2x(x-y)  +  (x  —  yy. 

34.  a2-4a(a-l)  +  4(a-l)^ 

35.  c2-6c(a-c)  +  9(a-c/. 

36.  m^  +  2  m  (m  —  n)  +  (wi  —  n)^- 

37.  16-24(a-6)+9(a-&)2. 

38.  a^  +  25 (y3 -  xf  -\-10x(f- x). 
39  14a(a;-t/)  +  (a;-2/)2  +  49a2. 
40.  10m(m  — 4)  +  25m*+(m  — 4)* 


FACTORING  96 

41.  {a-^hy-2{a  +  h){h-\-c)  +  {h  +  cf. 

42.  {a  -2  xf  +  A{a  -2x){2x  -h)  +  ^(2x  -b)\ 

43.  16(a  -  xf  +  32(a  -  x)  {x  +  h)  +  \Q{x  +  hf. 

44 .  (a  +  3  6)2  -  4  (a  +  3  6)  (3  6  -  2  c)  +  4  (3  &  -  2  c)". 

45.  {x"  +  X  +  iy  +2{x  +  l){ci?  +  X  +  1)  +  {x  +  If. 

46.  (a  +  &  +  c)2  +  2(a  +  6  —  c)  (a  +  6  +  c)  +  (a  +  6  -  c)». 

47.  (ar'  -  x'")''  +  2 (a;^  _  ^^^  (a;  +  1)  +  (a^  +  2  a;  +  1). 

128.   To  factor  the  difference  of  two  squares. 

1.  Since  {a-{-h){a  —  h)  =  a?~-h^,  what  are  the  factors  of  a?  —  V? 
How  do  these  two  factors  differ  ? 

2.  Since   (a^  +  6^  (a^  —  If)  =  a*  —  6^  what  are  the  factors   of 
a*  —b*?    How  do  these  two  factors  differ  ? 

Rule.  —  Find  the  square  roots  of  the  ttvo  terms,  mid  make  their 
sum  one  factor  and  their  difference  the  other. 

Sometimes  the  factors  of  a  number  may  themselves  be  factored. 

Examples 

1.  Factor  6^  -  y'^. 

Solution.  V^  —  y'^  ={b -\- y){h  —  y). 

2.  Factor  x^  —  1. 

Solution.  x^  —  \  ={x  +  \){x  —  \). 

3.  Factor  x*  —  1. 

Solution.  x*  -  1  =  (a;2  +  1)  (a;2  _  i) 

=:(x'!  +  l)(a;+l)(a;-l). 

Resolve  into  their  simplest  factors : 

4.  ar'-ml  9.   25 -c^.  14.  a^^-b\ 

5.  a^-y\  10.   a* -49.  15.  4a^-25t/l 

6.  a^-ie.  11.    x^-81.  16.  9  a^- 49  61 

7.  y^-a?.  12.    a* -16.  17.  a^x^-Ac?. 

8.  a2-9.  13.    a^-6*.  18.  m*  -  16  n*. 


96  FACTORING 

19.  25a:2_i  25  400a^-100t/l        31.  5a^-5. 

20.  81m*-l.  26.  2a^^2h\  32.  3a^-3a. 

21.  36  a* -25.  27.  4m*-4&*.  33.  x^-xf. 

22.  12162_c2.  28.  3  a;* -3/.  34.  5a^y-5ay. 

23.  400  a''- 81  yl  29.  5a;*-5y^".  35.  a^"  -  y^\ 

24.  lOOa^"  — 1.  30.  8x^^-Sf.  36.  cc2"+i  -  a;?/^". 

When  either  or  both   of  the  squares   are   polynomials,   the 
expression  may  be  factored  in  a  similar  manner. 

37.  Factor  25  a^  -  (3  a  +  2  bf. 

Solution 

One  factor  is  5  o  +  (3  a  +  2  6),  and  the  other  is  5  a  -  (3  a  +  2  6). 

5  a  +  (3  a  +  2  6)  =  5a  +  3a  +  2b  =  Sa  +  2b  =  2(4:  a  +  b). 

5 a  - (S  a  +  2  b)=  6  a  -  S  a  -  2  b  =  2  a  -  2  b  =  2(a  -  b). 

.-.  26  a2  -  (3  a  +  2  6)2  =  (5  a  +  3  a  +  2  6) (5  a  -  3  a  -  2  6) 

=  (8a  +  2b)(2a-2b) 

=  2(4a  +  6).2(a-6) 

=  4(4a  +  6)(a-6). 
Factor : 

38.  a^'-ia  +  bf.  42.  9  6^  _  (a  -  a;)* 

39.  b^-(2a+by.  43.  9  a^- (2  a -5)2. 

40.  a2_(5_^c)2  44^  a;*  -  (3  a^  -  2  2/)'^. 

41.  4c2-(6  +  c)2.  45.  49 a^- (5a -4 6)2. 

46.  Factor  (3  a  -  2  6)^- (2a -5  6)2. 

Solution 
(3a -2  6)2 -(2a -56)2 
=  [(3  a  -  2  6)  +  (2  a  -  5  6)][(3  a  -  2  6)  -  (2  a  -  5  6)] 
=  (3a-26  +  2a-56)(3a-26-2a  +  56) 
=  (5a-7  6)(a  +  36). 
Factor : 

47.  (2a  +  36)2-(a  +  6)2.  49.    (2  a;  +  5)^  -  (5  -  3  a;)^. 

48.  (5a -36)2 -(a -6)2.  50.    (a- 2  6)2  -  (a  -  5)2. 


FACTORING  97 

51.  {2x-Zyf-{^y  +  zy.  54.    {^x  +  &yf -{4.x-'6yf. 

52.  (5&-4c)2-(3a-2c)2.         55.    {a?  +  x^f  -  (2  x  +  2)\ 

53.  {A.x-^yy-{2x-^af.       56.    (a  +  6  +  c)'-(a-6 -c)^ 

57.  Factor  a^  +  4  —  c^  —  4  a. 

Solution 
a2  +  4  -  c2  -  4  a 
Arranging  terms,  =a2-4a  +  4-c2 

=  (a-2)2-c2 
=  (a-2  +  c)(a-2-c). 

58.  Factor  a2  +  62_c2_4_  2a6  + 4c. 

Solution 
a2  ^.  52  _  c2  _  4  -  2  a6  +  4  c 
Arranging  terms,  =  a^  -  2  a&  +  &^  -  c^  +  4  c  -  4 

=  (a2  -  2  a6  +  62)  _  (c2  _  4  c  +  4) 
=  (a-6)2-(c-2)2 
=  (a  -  &  +  c  -  2)  (a  -  6  -  c  +  2). 
Factor : 

59.  a^  -  2  ax  +  ar*  -  n^.  67.  (?  -  a" -h^ -2ab. 

60.  &2  4.2  6?/-j-2/2-w2.  68.  h''-x^-y'^  +  2xy. 

61.  l—4:q  +  Aq^  —  a^.  69.  ic^  —  x^  —  ^f —  2xy. 

62.  ?-2  -  2  ra;  +  a:^  -  16  «^  70.  9  c^  -  ar' - 1/^  +  2  a;y. 

63.  9a26-6a62+63_45c2  ^^^  a^  -  a^aj  -  4  ft^a;  -  4  a&a;. 

64.  4  a^c  +  12  a&c  + 9  6^0- 4  c».  72.  bc^  -  9  a^b  -  b^  -  6  ab^. 

65.  3a!2|/-12ay  +  12y3_3^^  73^  ab' -  4: a^  -  12  a'c  -  9 ac\ 

66.  4aw*-16aV+16a3-4an«.  74.  27  0^- 12a2  +  36  a&  -  27  6'' 

75.  a'-2ab  +  b^-c?  +  2cd-cP. 

76.  ar' -  2  a;?/ +  ^='-m2  + 10  m  — 25. 

77.  4a^  +  9 —  12  a; +  10  mn  —  m^  — 25  w^. 

78.  ar'  -  a2  +  ^'^  -  6^  +  2  xn/  -  2  a6. 

ALO.  — 7 


98  FACTORING 

129.  To  factor  a  quadratic  trinomial. 

(a;  +  3)(a;+  5)  =  o^  +  8  a; +  15. 
{x-S){x-  5)  =  x'-  8a; +  15. 
(«  + 1)  (a;  + 15)  =  a^  +  16  a;  +  15. 
(a;  -  1)  (x  +  15)  =  a^  +  14  a;  -  15. 
(a;-3)(a;+  5)  =  a^+  2a; -15. 
(a; +  3) (a;-    5)  =  ay' -    2 a; -15. 

1.  How  may  the  first  term  of  each  factor  be  found  from  the 
product  ? 

2.  When  the  last  term  of  the  product  has  the  sign  +,  how  do 
the  signs  of  the  last  terms  of  the  factors  compare  ?  How,  when 
the  last  term  of  the  product  has  the  sign  —  ? 

3.  How  does  the  coefficient  of  the  middle  term  of  the  product 
compare  with  the  algebraic  sum  of  the  last  terms  of  the  factors  ? 

130.  A  trinomial  of  the  form  cn^  -{-bx  +  c,  in  which  x^  is  the 
square  of  any  number,  c  the  product  of  two  numbers,  and  b  their 
algebraic  sum,  b  and  c  being  either  positive  or  negative,  is  called 
a  Quadratic  Trinomial. 

Rule.  —  Arrange  the  trinomial  according  to  the  descending  jwivers 
of  one  of  the  letters. 

For  the  first  term  of  each  factor  take  the  square  root  of  the  first 
term  of  the  trinomial;  and  for  the  second  terms,  such  numbers  that 
their  product  is  the  third  term  of  the  trinomial,  and  their  algebraic 
sum  multiplied  by  the  first  term  of  the  factor  loill  be  equal  to  the 
second  term. 

Examples 

1.    Resolve  a^  —  13  a;  —  48  into  two  binomial  factors. 

Solution.  —  The  first  term  of  each  factor  is  evidently  x. 
Since  the  product  of  the  second  terms  of  the  two  binomial  factors  is  —  48, 
the  second  terms  must  have  opposite  signs ;  and  since  their  algebraic  sum, 

—  13',  is  negative,  the  negative  term  must  be  numerically  larger  than  the 
positive  term. 

The  two  factors  of  —  48  whose  sum  is  negative  may  be  1  and  —  48,  2  and 

—  24,  .3  and  —  16,  4  and  —  12,  or  6  and  —  8.     Since  the  algebraic  sum  of  3 
and  —  16  is  —  13,  3  and  —  16  are  the  factors  of  —  48  sought. 

.-.  x2  _  i3x  _  48  =(a;  +  3)(a:  -  16). 


FACTORING  99 

2.  Factor  rn?  +  m  —  72. 

Solution.  —  Since  +  9  and  —  8  are  the  only  two  factors  of  —  72  whose 
algebraic  sum  is  +  1,  the  coefficient  of  the  middle  term,  the  required  factors 
are  (m  +  9)  and  (m  —  8). 

.-.  r/i2  +  m  -  72  =(w  +  9)(m  -  8). 

Separate  the  following  into  their  simplest  factors : 

3.  x^  +  7x  +  12.  18.    a^  +  5aa;  +  6al 

4.  ;y2_7  2/  +  12.  19.    o?-&ax-\-ba?. 

5.  p2_8^_^12.  20.   y^  -  i  by- 12  b\ 

6.  ,-2  4.8r  +  12.  21.    /-3wy-28ji2. 

7 .  m^  +  5  m  — 14.  22.2;^  —  anz  —  2  a-wl 

8.  a^-2a-15.  23.    «*  + 19 cx^-f 90 c^. 

9.  6-  +  &-12.  24.    »«  + 12  aa^ -I- 20  a". 

10.  »^  +  r-30.  25.  a;^«-lliV  +  24  6*. 

11.  c^  —  c  — 72.  26.  5  wa;''' —  55  wa:  + 150  n. 

12.  c2_5c-14.  27.  3tt26aj2_3a2to-6a26. 

13.  x2_a;-110.  28.  12  m^a;^  -  60  m^fea;  +  72  m^^^. 

14.  a2  +  9a-52.  29.  4 aa;  +  2 aa;- - 48 a. 

15.  a2  +  8a-128.  30.  11  a-;«- 55 aa;  + 66a;. 

16.  ar^- 25a; +  100.  31.  20  6a;  +  10  6^  -  630 a;l 

17.  ar'  + 12  a; -85.  32.  x^ -{-(b  —  a)x  ~  ab. 
131.    To  factor  trinomials  of  the  form  ax^  +  bx  -\-  c. 

Examples 
1.    Separate  3  a;^  +  11  a;  +  6  into  two  binomial  factors. 

Solution.  —  Since  3  x^  is  the  product  of  the  first  terms  of  the  factors,  and 
6  is  the  product  of  their  last  terms,  and  since  the  only  factors,  each  contain- 
ing X,  that  3  x2  can  have  are  3  x  and  x,  one  factor  is  3  x  plus  a  factor  of  6,  and 
the  other  is  x  plus  the  other  factor  of  6.  By  trial,  it  is  found  that  2  and  3 
are  the  factors  of  6,  and  that  if  2  is  added  to  3  x  and  3  to  x,  the  middle  term 
of  the  product  (3  x  +  2)  (x  +  3)  will  be  11  x. 

.-.  3x2+ llx  +  6=:(3x  +  2)(x  +  3). 


100  FACTORING 

2.   Factor  9«2  +  30a;  +  16. 

Solution 

9  a;2  +  30  a;  +  16 
=  (3a;)2+10(3x)+  16 
Put  m  for  3  r,  =  m^  +  10  m  +  16 

§  130,  =  (m  +  2)  (m  +  8) 

Put  3  a;  form,  =(3x  +  2)(3x  +  8). 

Suggestion.  —  When  the  coefficient  of  x^  is  a  square,  and  when  the  square 
root  of  the  coefficient  of  yfi  is  exactly  contained  in  the  coefficient  of  x,  the 
trinomial  may  be  factored  by  the  method  illustrated  above. 


3.    Factor  4a^  — 5x  — 6. 


Solution 


4x2  -  5x -6  =(4x2  -  5x  -  6)  X  *  =  1^^^^^20x^^ 
^  ^4  4 

_  (4x)2  -  5(4  x)  -  24  ^  (4  X  -  8)(4  X  +  3) 
4     ■  4 

=  ^(^-Y^  +  ^)=(x-2)(4x  +  3). 

Explanation.  —  Although  the  first  term  is  a  square,  its  square  root  is  not 
exactly  contained  in  the  second  term. 

But  if  such  a  trinomial  is  multiplied  by  the  coefficient  of  x^,  the  resulting 
trinomial  will  be  one  whose  second  term  exactly  contains  the  square  root  of 
its  first  term. 

Multiplying  the  given  trinomial  by  4,  factoring  as  in  example  2,  and  divid- 
ing the  result  by  4,  the  factors  of  the  given  trinomial  are  (x— 2)  and  (4  x+3). 

4.   Factor  24  x^  +  14.  x  -  5. 

Solution 

24x2-|-14x-5=(24x2+  14 x  -  5)  x  -  =  ^M^Jl^I^^iJO 

6  6 

_  (12  x)2  +  7(12  X)  -  30  _  (12  X  +  10)  (12  x  -  3) 
6  6 

=  ^(^^  +  t^-^i^^-l)=(6x  +  5)(4x-l). 

A  X  o 

Suggestion.  — When  the  first  term  is  not  a  square,  it  may  always  be  made 
&  square  whose  square  root  will  be  exactly  contained  in  the  second  term  by 
multiplying  the  trinomial  by  the  coefficient  of  x^. 


FACfORmG  101 

After  factoring,  the  result  should  be  divided  by  the  coefficient  of  x^. 
Frequently  the  multiplier  can  be  taken  smaller  than  the  coefficient  of  Z'^, 
as  in  the  above  example. 

Separate  into  their  simplest  factors  : 

5.  2x'  +  x-15.  17.  9 a?*  -  10 or*  -  16. 

6.  9  ^^-42  a; +  40.  18.  27&*-3  62_l4. 

7.  5a^  +  1.3a;  +  6.  19.  10  a;«  -  2  a.^  -  44. 

8.  ^x^-llx  +  lQ.  20.  2a?  +  5xy  +  2j^. 

9.  25ar'  +  15a;  +  2.  21.  20^= +  3a;^  -  2/. 

10.  16a^  +  20a;-66.  22.  3 a^  -  10 a;?/ +  3  j/^. 

11.  36 x-2  -  48 a;  -  20.  23.  6a^-lla;-35. 

12.  9a;- +  43a;- 10.  24.  6 a;^  _  ^3 3.  _,_  g 

13.  25a;2^25a;-24.  25.  15  a;^  -  14 a;  -  8. 

14.  49a;2-42a;-55.  26.  Wx^  +  llx-i:. 

15.  16x-2  +  50a;-21.  27.  21  a^- a- 10. 

16.  4a.-2-4a;-35.  28.  18  ar^  -  3 a;  -  36. 

132.   To  factor  the  sum  of  the  same  odd  powers  of  two  numbers. 

Examples 

1.  Factor  dJ  +  h^. 

Solution 

§§  111-113,  a^  +  6^  =(a  +  6)  (06  -  a^b  +  a^h"^  -  a^h^  +  a^ft*  -  ah^  +  b^). 

2.  Factor  m''  +  32ar*. 

SOLDTION 

m^  +  32  a;5  =  ms  +  (2  a;)^ 
§§  111-113,  =  (m  +  2  a;)(m*  -  2  mH  +  4  rrfir?  -  8  mx^  +  16  x*). 

3.  Factor  »*  +  /. 

Solution 

X«  +  2/«=(x2)8+(y2)8 
§§    111-113,  =  (X2  +  2/2)  (X*  -  X2y2  +  2/4). 


102  FACTORING 

Factor  the  following : 

4.  m^  +  n^.  7.  ar*  +  l.  10.  a"  +  32. 

5.  a?  +  x\  8.  r^  +  s^  11.  p^  +  21. 

6.  a^  +  ft'*.  9.  x>-{-y^.  12.  a;"  +  128. 

133.  To  factor  the  difference  of   the  same   odd   powers   of   two 
numbers. 

Examples 

1.  Factor  a^  —  &^ 

Solution 

§§  111-113,  a'  -W  ={a-  b)(a^  +  a^b  +  a*b^  +  a^b^  +  a'^b*  +  a¥  +  b^). 

2.  Factor  m^  -  32  rc^. 

Solution 

§§  111-113,  wi5  -  32  x8  =  m5  -  (2  «;)6 

=  (m  -  2  X)  (m*  +  2  ??i3a;  +  4  ,^23:2  +  8  nix^  +  16  x*). 

3.  Factor  a'"  — 6^ 

Solution 

§§    111-113,     alO  -  65   =(a2)5_  65=,(a2  _  ft)(«8  +  ^65  +  ^452  +  ^263  +  64), 

Factor  the  following : 

4.  a'-6^  7.  1/-8.  10.  ar^ -?/'". 

5.  a^-6''.  8.  a^-32.  11.  a.-«  -  ?/3 

6.  ?/^-al  9.  a^-l.  12.  ar'-243. 

134.  To  factor  the  difference  of  the  same  even  powers  of  two 
numbers. 

Examples 
1.    Factor  a^  —  b^ 

First  Solution 

§§  111-113,  a6  -  &6  =(a  -  b)(a^  +  a^b  +  a^b^  +  a'^b^  +  ab*  +  b^) 
=  (a-b)  {cfi  +  aH)^  +  a*6  +  a6<  +  a%'^  +  b^) 
=  {a-  b)\_d\a^  4-  fts)  +  ab{a^  +  &3)+  62(^3  +  js)] 
=  (a  -  6)  (a2  +  aft  +  62)  (a8  +  63) 

§  132,  =  (a  -  6)  (a2  +  a6  +  62)  («  +  6)  {cfi  -  a6  +  62). 


FACTORING  103 

Second  Solution 
§  128,  a6  -  66  =  (a3)2  _  (63)2  ^  (^ta  ^  ^3)  (^8  _  ^S) 

§§  132,  133,  =  (a  +  6)  (a^  -  a6  +  &2)  (a  _  ft)  (a2  +  a&  +  ft^)- 

In  each  of  the  above  solutions  a^  —  h^  is  first  separated  into  two  factors. 
Thus,  a^  —  b^  is  divisible  by  a  — 6  ;  also  by  a^— 6^,  since  a**  — 66=(a')2— (ft*)2. 

When  the  even  powers  are  regarded  as  squares,  as  in  the  second  solution, 
the  process  may  be  regarded  &&  factoring  the  difference  of  two  squares. 

Separate  the  following  into  their  simplest  factors : 


2. 

x"  — /. 

5.    x*-16. 

8. 

l-6«. 

3. 

x'  -  1. 

6.    a;* -81. 

9. 

64-/. 

4. 

a«  -  &«. 

7.    a' -625. 

10. 

1-^8. 

Fac 

itor  the  following  by  previous  principles : 

11. 

y-q\ 

17.    ?'«-s«. 

23. 

1+a^. 

12. 

p^-h<f. 

18.    /  +  s«. 

24. 

X  +  2C^. 

13. 

p"  -  ^. 

19.    m^  —  n^ 

25. 

32  n  -  n« 

14. 

r*  -  s*. 

20.    m'  +  w^ 

26. 

m^  —  a^. 

16. 

r'-^. 

21.    m^  —  n^ 

27. 

6^  -  a^ft*. 

16. 

1^  +  ^. 

22.    a«-l. 

28. 

125  +  a\ 

135.   To  factor  by  the  Factor  Theorem. 

1.  If  5  aj  =  0,  what  must  be  the  value  of  05  ? 

2.  If  5  (a;  —  3)  =  0,  which  factor  reduces  the  first  member  to 
zero ;  that  is,  what  vabie  of  x  reduces  it  to  zero  ? 

3.  Since  the  expression  5 (a;—  3)  reduces  to  zero  when  a;  —  3 
reduces  to  zero,  and  a;  —  3  is  reduced  to  zero  by  substituting  3 
for  X,  what  value  substituted  for  x  will  reduce  to  zero  every  expres- 
sion of  which  a;  —  3  is  a  factor  ? 

4.  How,  then,  may  it  be  discovered  whether  a;  —  3  is  a  factor 
of  a  given  expression  containing  x  ? 

5.  What  value  of  x  will  reduce  the  expression  a.*'  —  8  to  zero  ? 
What  factor,  then,  has  a^  —  8  ? 

6.  What  value  of  x  will  reduce  the  expression  a^  —  8  a;  +  7  to 
zero  ?    What  factor,  then,  has  a^-^  —  8  a;  4-  7  ? 


104  FA  CTORING 

7.  When  does  a  rational  integral  expression  containing  x  have 
the  factor  x-1?  x-2?  x-S?   x-a? 

136.  Factor  Theorem.  —  If  a  rational  integral  expression  containing 
X  redvbces  to  zero  when  a  is  substituted  for  x,  it  is  exactly  divisible 
by  x  —  a. 

Demonstration.  — Let  D  represent  any  rational  integral  expression  con- 
taining X,  and  let  D  reduce  to  zero  when  a  is  substituted  for  x. 

It  is  to  be  proved  tliat  D  is  exactly  divisible  by  x  —  a. 

Suppose  that  the  dividend  D  is  divided  by  x  —  a  until  the  remainder  does 
not  contain  x.     Denote  the  remainder  by  B  and  the  quotient  by  Q. 

Then,  D  =  Q(x  -  a)+  B.  (1) 

But,  since,  by  supposition,  D  reduces  to  zero  when  x  =  a;  that  is,  when 
X  —  a  =  0,  equation  (1)  becomes 

0  =  0+  R; 
whence,  ^  =  0. 

That  is,  the  remainder  is  zero,  and  the  division  is  exact. 

Note,  a  is  the  known  number  substituted  for  x,  and  it  may  be  either 
positive  or  negative. 

Examples 

1.  Factor  a;'  —  8  x^  4- 17  ic  —  10  by  the  factor  theorem. 

Solution 

jc3  _  8  a;2  +  17  X  -  10 

=  (a;-l)(x2_7a;+10) 

=  (a;-l)(a;-2)(x-5). 

Since,  when  1  is  substituted  for  x,  x^  —  Sx^  +  17x  —  10  reduces  to  0, 
X  —  1  is  a  factor  of  the  expression.  Dividing  by  x  —  1,  the  other  factor  is 
x^  —  7  X  +  10,  whose  factors,  §  130,  are  (x  —  2)  and  (x  —  5). 

2.  Factor  a^  +  a^  —  16a^  —  4a;-|-48  by  the  factor  theorem. 

Solution 

X*  +  x3  -  16  x2  -  4  X  +  48 
=  (X  -  2)(x8  +  3  x2  -  10  X  -  24) 
=  (x  -  2) (x  +  2)(x2  +  X  -  12) 
=  (x-2)(x4-2)(x-3)(x  +  4). 

Since,  when  2  is  substituted  for  x,  the  expression  reduces  to  0,  x  —  2  is 
a  factor  of  the  expression.  Dividing  by  x  —  2,  the  other  factor  is  x^  +  3  x^ 
—  lOx  -  24.  Since,  when  —  2  is  substituted  for  x,  x^  +  3x*  —  lOx  —  24 
reduces  to  0,  x  +  2  is  a  factor  of  x'  +  3  x^  —  10  x  —  24. 


FACTORING  105 

Dividing  by  x  +  2,  the  other  factor  is  oc^  +  x  -  12,  a  quadratic  trinomial 
whose  factors,  §  130,  are  (a;  -  3)  and  (x  +  4).  Hence,  the  factors  of  the 
given  expression  are  («  -  2),  (x  +  2),  (x  -  3),  and  (x  +  4). 

Factor  the  following  polynomials  by  the  factor  theorem  : 

3.  x'-dlx  +  dO.  21.  y?-lx  +  Q. 

4.  4a;2-7a;  +  3.  22.  a^- 19  a; +  30. 

5.  26a^-10a;-16.  23.  a.-^ - 67 a; -  126. 

6.  48  a^  -  31  a;  -  17.  24.  a^- 39  a; -70. 

7.  36  a^  -  61  a;  +  25.  25.  a'' +  4a2- 11a -30. 

8.  a^- 9x2  + 23a; -15.  26.  a'^ +  9a2  +  26a +  24. 

9.  3^-13x2  + 47  a; -35.  27.  m'^  -  6m2  -  m  +  30. 

10.  a^-14a:2  +  35a;-22.  28.  6^  -  5  6^ -29&  +  105. 

11.  0^-4^2 -7a; +  10.  29.  a'^  +  lOa^  -  17a- 66. 

12.  a;3-6a;2-9a;  +  14.  30.  m='  + 7m2  + 2m  -  40. 

13.  a;^  -  12  a;2  ^  41  ^  _  30.  31.  &3_^  16  6^  +  736  + 90, 

14.  .r' -  11  a;' +  31  a;  -  21.  32.  «»  + 12 11^  +  41  n +  42. 

15.  ar^-10a;2^29a;-20.  33.  a;*  -  15  ar' + 10  a;  +  24. 

16.  a;3- 16  3^^  +  71  a; -56.  34.  a;*  -  25  a;^  ^  60  ^. ._  36 

17.  a;' -57  a; +  56.  35.  a;*  +  13 ar^  -  54  a;  +  40. 

18.  a;^- 21  a; +  20.  36.  a;*  +  22  a;^  +  27  a;  -  50. 

19.  ar'- 31  a; -30.  37.  x^  -  ^  x^y""  -  4:  xf  +  12 1/. 

20.  ar'- 13  a; +  12.  38.  x* -'da?y'' +  12xif  -  4.y^. 

39.  a;*-a;»-7ar'  +  a;  +  6. 

40.  a;^  -  9  a;3  + 213^^  +  0; -30. 

41.  a;*  +  8  ar^ +  14  3^^-8  a; -15. 

42.  a;^-4a;^  +  19a;--28a;  +  12. 

43.  a;«-18a^  +  30.'»2-19a;  +  30. 

44.  a;«-10a;*  +  40a;3_30ar'+80a;-32. 


106  FACTORING 

SPECIAL  APPLICATIONS  AND  DEVICES 

Examples 

137.  1.   Factor 

a^  +  W  +  c^  +  d^  +  2ah  -2  ac+2ad  -2bc  +  2bd -2cd. 

Solution.  —  Since  the  polynomial  consists  of  the  squares  of  four  inirabers 
together  with  twice  the  product  of  each  of  them  by  each  succeeding  number, 
the  polynomial  is  the  square  of  the  sum  of  four  numbers,  §  95,  and  may  be 
separated  into  two  equal  factors  containing  a,  6,  c,  and  d  with  proper  signs. 

Since  —  2  ac,  —  2  be,  and  —2  cd,  the  products  that  contain  c,  are  negative, 
while  2  ab,  2  ad,  and  2  bd,  the  products  that  do  not  contain  c,  are  positive, 
it  is  evident  that  the  sign  of  c  is  the  opposite  of  that  of  a,  b,  and  d. 

Therefore,  the  factors  are  either 

{a  +  b  —  c  +  d)  (a  +  b  —  c  +  d), 
of  (—  a  —  b  +  c  —  d)(—  a  —  b  +  c  —  d). 

Factor  the  following : 

2.  9x^  +  4:y-  +  25z^-12xy  +  30xz-20yz. 

3.  25  m^  -\- 36  n^  •{- p^  —  60  mn  —  10  7np  +  12  np. 

4.  a"  -^16  x'  +  S6  if  -Sax"  +  12  ay  -4.8  x^y. 

5.  x'  +  ia^  +  b^  -\-  y^  -\-  Aax  —  2  bx  +  2  xy  ~  4:ab  +  4:ay  -  2by. 

6.  m^  +  4  ?i^  +  a^  +  9  —  4  mn  —  2  am  +  6  m  +  4  an  —  12  )i  —  6  a. 

138.  The  principle  by  which  the  difference  of  two  squares  is 
factored  has  many  special  applications. 

1.  Factor  a*  +  a^b^  +  b\ 

Solution.  —  Since  «■*  +  a^b"^  +  b*  lacks  +  a^^a  of  being  a  perfect  square, 
and  since  the  value  of  the  polynomial  will  not  be  changed  by  adding  a'^6* 
and  also  subtracting  a^b'^,  the  polynomial  may  be  written 

a*  +  2  a262  +  ^4  _  ^52^2^ 
which  is  the  difference  of  two  squares. 

a*  +  a'^b^  +  6*  =  a*  +  2  a%'^  +  b*  -  a%^ 
=  {a^  +  b^)2-(aby 
=  (a2  +  ab  +  b-^)  (a^  -  ab  +  ¥). 

2.  Factor  4  a* -13  a- +  9. 

Solution.  4  a*  -  13  a'^  +  9  -  4  a*  -  12  a"^  +  9  -  cfi 

=  (2  a2  -  3)2  -  «2 
=  (2  a-  +. a  -  3)  (2  a-  -  a  -  3). 


FACTORING  107  - 

3.  Factor  a*  +  4. 

Solution.  a*  +  4  =  a*  +  4  a^  +  4  -  4  a^ 

=  (a^  +  2)2  -  (2  ay- 
=  (a2  +  2a  +  2)(a2-2a  +  2). 

Factor  the  following : 

4.  X* -\- x'y'' +  y\  11.    a;*  +  a^  +  l. 

5.  a«  +  a*6*  +  &^.  12.    7t«  +  ?i*  +  l. 

6.  p*+pY  +  <1*-  13.    16a;*  +  4a.V  +  ?/^. 

7.  9  a^  +  20  ar'/  + 16  2/*.  14.   a^6*  -  21  a^d^  +  36. 

8.  4  a*  +  11  a^fe^  +  9  6*.  15.    c*  +  c^d V  +  c? V. 

9.  16a^-17aV  +  a;^  16.   25  a^  -  14  a^ft*  +  61 

10.  25 a;* -29 0^2/2  +  4?/*.  17.   9  a*  +  26  a^ft^  +  25 5*.      ' 

18.  &^  +  64.  21.  a* +  324.  24.   x^  +  Uy\ 

19.  a* +  4 6*.  22.  a«-16.  25.    4 a* +  81. 

20.  m^  +  4.  23.  nv"  +  4  mn*.  26.    a^/  +  4  xy^. 

139.   Many  polynomials  may  be  written  as  quadratic  trinomials 
in  which  o?  and  x  are  replaced  by  polynomials. 

1.    Factor  9^^  ^  4^2  ^  1222  +  21  icz  +  14?/z  +  i2xy. 

Solution.  9  x^  +  4  2/2  +  12  2;2  +  2I  a;g  +  14  yz  +  12  xy 

=  (9a;2  +  12x?/  +  4j/2)  +  (21a;2  +  14t/0)+1222 
=  (3  a;  +  2  j/)2  +  7  2(3  X  +  2  y)  +  4  2  .  3  « 

§130,  =(.Tx  +  2j/  +  4«)(3x  +  22/ +  3«). 

Factor  the  following : 

2.  a2  +  2 a&  +  &2  +  8 ac  + 86c  +  15 c2. 

3.  0^2  _  6  a;?/  +  9  /  +  6  a;z  —  18  2/2  +  5 z^. 

4.  m?  +  n^  —  2mn  -\-l mp  —1  np  —  30 p*. 

5.  16  n^  +  55  -  64  w  -  16  m  +  m2  + 8  wn. 

6.  9m*  +  fc2  _  30  +  39m''  +  13  A;  +  6 m?k. 

7.  25a2  +  2/2  +  10x2  +  10a?/-35ax-7a^. 

8.  4  a;2  _|_  ^2  _  g  2,2  _  4  ^.^^  4.  2  a;2!  —  2/2. 

9.  a2  +  62  +  c2  +  2a6  +  2ac  +  26c  +  5a  +  56  +  5c  f  6. 


108  FACTORING 

REVIEW  OP  FACTORING 
140.    Factor  the  following  : 

1.  ire-n\  11.  p*  +  4:.  21.   4a;*-4aj. 

2.  x^-l.  12.  1  +  0^2.  22.    7?/* -175. 

3.  y*-l.  13.  y-a*y.  23.    8-27aV. 

4.  1-0^.  14.  x'y-f.  24.    32a; -2a^. 

5.  ajio-l.  15.  a^3_^5i2  25.    6  &*  +  24. 

6.  x^-1.  16.  a* -256.  26.   a* +  27  a-. 

7.  l-a^.  17.  3a3-3a.  27.    ft^  -  196. 

8.  1-6*.  18.  64-2?/^.  28.    450 -2a2. 

9.  a-dJ.                   19.  7?i^-7n.  29.    4m3  +  .004. 
10.    H' +  h.                   20.  a^-9a.  30.    125-8iB«. 

31.  ic2_25a;  +  100.  45.  y'^-xy-42x'. 

32.  x'-xy-lS^yK  46.  ar*  -  aa;  -  72  a^. 

33.  aa^-3aa;  — 4a.  47.  n^—an  —  ^Oo?. 

34.  ar''  +  5a^-6a;.  48.  a%'' -\- ah  -  5%. 

35.  3ar'  +  30.^  +  27.  49.  10  a2c  + 33  ac  -  7c. 

36.  128  a^- 250  a^  50.  60w/-61  wy- 56n. 

37.  5a;i«  +  10af-15.  51.  25  a^  +  60  a^  +  36 1/^ 

38.  6  .^2  -  19  a;  +  15.  52.  Q  ay?  +  5  axy  -  (^  ay\ 

39.  a-2"  +  2af3/^  +  /^.  53.  169  a^  -  26  aaf  +  aV. 

40.  7:i?-nxy-Mf.  54.  aV  +  a^&V+ftl 

41.  /-25?/a;  +  136a«.  55.  16  a;*  +  4  a^i/^  +  y*. 

42.  9ar^-24a;?/  +  16?/2.  56.  6*0-13  6^0  +  420. 

43.  289  a^- 34  an/ +  2/2  57  2  a'' -  6  a6  -  140  6=*. 

44.  3  6a^  +  6a;2/ -  10  6?/^.  58.  m«»  -  21  mn^  +  80  n». 


FACTORING  109 

59.  17  0^  +  25.^-18.  74.  x'  +  x'y-^lxif-lObf. 

60.  5x^-2Qxy  +  by\  75.  a? -ex -\-2dx-2cd. 

61.  y2  _j_  16 ay  -  36 a*.  76.  o^y  +  ^x^y -^Ixy —  Idy. 

62.  8a2-21a6-962.  77.  a^- 3aa;  +  4te-12a&. 

63.  60a^  +  8aa;-3a*.  78.  oar' - 9 ar=  +  26 aa; - 24 a. 

64.  30  ar' -  37  a;  -  77.  79.  12aa;-8&a;-9ai/  +  6  6y. 

65.  2a^  +  28a,-2  +  66a;.  80.  25  a^  -  9  y^  _  24  ?/2  - 16 2^. 

66.  a2  +  62_c2_2a&.  81.  x'-z''  +  y''-a''-2xy  +  2az. 

67.  aar'  +  10ax-39a.  82.  2h'^m-3aW-\-2hmx-^abx. 

68.  n*  +  wW6*  +  a*6»  83.  0^+1^+0^-2  ah -2 ac-\-2  he. 

69.  aV  +  aV  +  al  84.  a^t/ + 14  a^y  +  43  a^  +  30  y. 

70.  a2-16a-17.  85.  o^y - 15 ar'?/ +  38 a^  - 24 y. 

71 .  a^x^  —  4  aaj  +  3.  86.  aha?  +  3  aha?  —  ahx  —  3  ah. 

72.  &8  +  6y  +  2/^  87.  3&ma;  +  2&m  — 3ana;— 2aM. 

73.  a;^-2a;«  +  a;.  88.  20  aa^ - 28  aa^+ 5  a^x- 7  al 

89.  x'  +  ^f  +  26z^-Qxy-Wxz  +  ^0yz. 

90.  9ar'  +  2/^  +  16z2_6a;2/-8.V2  +  242a;. 

91.  a?yh^  +  a-h^  +  l  +  2ahmz  +  2xyz  +  2ah. 

92.  a^"^  +  h^(?  +  <?d?  -  2  ahh  +  2  ahcd  --  2  he'd. 

93.  a^  +  nV  +  w»  +  27iV  +  2nV  +  2wV. 

94.  a^Wa?  -  a^ft*  -  6V  +  h""  -  a^a?  +  a"  +  a?  - 1. 

95.  (aH-6)«-l.  100.  3a^  +  96a;. 

96.  a^- 2  0^  +  1.  101.  (a -2)'^+ (a -1)1 

97.  &3_452_,_8.  102.  12a^  +  3a^-8a;-2. 

98.  a^-10ar'  +  125.  103.  2  a:^  _,_  ^q  a;  +  aa;  +  5  a. 

99.  8a;*-6ar'-35.  104.  a.-^  +  5  ar^  -  29  a;  -  105. 

105.   7n?n^  +  a^h^  +  lM  +  2hmv?  +  2ah''n  +  2abmn. 


110  FACTORING 

106.  a*  +  a%  +  a^h""  +  arb^  +  ah*  +  &^ 

107.  a262_4a6a;-4cc  +  2a&  +  4a^. 

108.  (a+6)='(a;-y)-(a  +  &)(a^-y2). 

109.  1  —  a^  +  abx'  +  ^a;''  —  6a;  —  a&. 

110.  31?  —  2(? -{•  x^y  —  xy -\- 2(?y  —  xy^. 

111.  a^"-2  +  6y  +  2  a;"-i&v. 

112.  :i?  +  Ws^  +  15x  +  12b. 

113.  4  (a6  -I-  cdf  -  (a^  -|-  52  _  ^2  _  ^2^2^ 

114.  x^  —  a^.  125.  a;^  +  4a:. 

115.  {a^  +  ¥-cy-4.o?b\  126.  x^  -  ar^  -  a;*  +  a;3_ 

116.  a*62  + 0,26-12.  127.  (a  +  hy-(h-c)\ 

117.  x^  —  xy  —  x^y  +  y^.  128.  3 «&(«  +  &)  + a'' +  6^ 

118.  .T*  -  4  x-/ f  2  x^  -  16  ia''.  129.  (x  +  ^)3  +  (a;  -  2/)^ 

119.  a*-h*-{a  +  h){a-h).  130.  a^' -  (a  +  Syi 

120.  x^- 6^2 +  12 a; -8.  131.  x'-l\^xY  +  y*- 

121.  1000  x^  -  27  ?/3.  132.  vv"  +  m- -  mn  -  myi\ 

122.  (a  +  a;)*  — a;^  133.  (a?  —  y^  —  (a?  —  xyf. 

123.  l+(a;+l)3.  134.  x^  -  f  -  3  a^y^x- -  y^). 

124.  ab  -  bx"  +  x^y'"  —  ay"'.  135.  (x^+6x+9y-(x^+ox+6y. 

136.  ar^  +  (a  +  6  —  c)a^  +  (0,6  —  ac  —  6c) .t  —  a6c. 

137.  Factor  32  —  a;^  by  the  factor  theorem. 

138.  Factor  16  +  5  x  —  11  ar'  by  the  factor  theorem. 

139.  If  n  is  odd,  factor  a"  —  «"  by  the  factor  theorem. 

140.  If  n.  is  odd,  factor  x"  +  ?•"  by  the  factor  theorem. 

141.  Factor  a?  —  6  bx^  +  12  b'\v  —  8  6^  by  the  factor  theorem. 

142.  Discover  by   the  factor  theorem  for  what  values   of  n, 
between  1  and  20,  x"  +  «"  has  no  binomial  factors. 


FACTORING  111 

EQUATIONS   SOLVED   BY   FACTORING 
141.    1.   Find  the  value  of  a;  in  o^  +  1  =  10. 


a^  +  1  =  10 


FIRST  PROCESS  EXPLANATION.  —  Transposing    the 

known  term  1  to  the  second  member, 
the  first  member  contains  the  second 
x*  =    9  power,  only,  of  the  unknown  number. 

X'X  =iZ  '3  .*.  a;  =  3  Separating  each  member  into  two  equal 

factors, 


a.a:  =  3.3    ora!;-x  =  — 3-  —  3. 


or  a;  •  a;  =  —3  •  —3  .-.  x  =  —3 

.-.  a;  =  ±  3 

Since,   if  a;  =  3,   x  •  a;  =  3  •  3,    and   if 

a;  =  —  3,  a;-x  =  —  3-—  3,  the  value  of  x  that  makes  x'^  =  9,  or  that  makes 

x2  4-  1  =  10,  is  either  +  3  or  -  3  ;  that  is,  x  =  ±  3. 

Find  the  values  of  x  in  the  following  equations : 

2.  a^  +  3  =  28.  7.    a^- 24  =  120. 

3.  ar^  +  l  =  50.  8.    ar^  +  ll  =  180. 

4.  x2-5  =  59.  9.    ar^-ll  =  110. 

5.  a^-7  =  29.  10.    a^-62=a2-2a6. 

6.  a:^  +  3  =  84.  11.    ar*  — 4  w^  =  m'^  —  4mw. 

12.  Find  the  value  of  a;  in  a^  +  1  =  10. 

Explanation. — The  first  process 
SECOND   PROCESS  is  given  in  example  1. 

o  ,   H  H  rv  In  the  second  process,  all  terms 

are    brought    to    the    first  member, 
^  —  9  =    U  which  is  factored  as  the  difference  of 

/jp  _  3"\  (a;  -f  3)  =    0  the  squares  of  two  numbers. 

n      rx       ^  o  Since  the  product  of  the  two  fac- 

••.   a;  —  3  =  0,  whence  x  =  6  .       .    ^  *  *,       •    i  +^  n 

'  tors  IS  0,  one  of  them  is  equal  to  0. 

or  a; +  3  =  0,  whence  a;  =  —3       This  gives  x-3  =  0  or  x  +  3  =  0; 
•    a;  =  +  3  whence  x=:3  or  x  =  —  3j   that  is, 

X  =  ±  3. 

Solve  the  following  equations ; 

13.  x2  +  35  =  39.  16.  a^-31*=0. 

14.  a^-50  =  50.  17.  a^- 4  6^  =  0. 

15.  a^  +  90  =  91.  18.  ar^-9w2  =  0. 


112  FACTORING 

19.  or' -21  =4.  24.  .32-0^  =  28. 

20.  ar^-56  =  8.  25.  65  -  a:^  ^  jg 

21.  a^-3a2  =  6a*.  26.  40^! - 862=  Sd^. 

22.  x'-\-bh*  =  Qh\  27.  ar' +  25  =  25  +  m". 

23.  x2_4o  =  24.  28.  ar' -  30  =  2  (2  6'^  -  15). 

29.  Solve  a^  +  2  atn  =  a^  -\-  m^. 

Solution 

x^  +  2  am  =  a"^  +  in^. 

x2  =  a2  _  2  am  +  m*. 
X  •  05  =  (a  —  jw)  (a  —  m) , 
or  X  •  X  =  —  («  —  m)  •  —  (a  —  m). 

.•.  X  =  ±(a  —  m). 

Solve  the  following  equations : 

30.  a^-c2  =  d2-2cd.  36.  ar^  -  c^  =  36  -  12  c. 

31.  a;2-62  =  46c  +  4c2.  37.  a^  -  4  6^  =  36  -  24&. 

32.  a^-w2  =  6w  +  9.  38.  aj2-a2  =  9 -6a. 

33.  a^  +  10a  =  a2  +  25.  39.  a,'^  -  &^  =  4 -4ft2. 

34.  a?-a?  =  2a  +  l.  40.  a^  -  a^ft"  =  2a&  +  1. 

35.  ar'-m2  =  8m  +  16.  41.  a?  -  r*  =^  h*  -  2  7^h\ 

42.    Find  the  value  of  a;  in  a^  +  4  a;  =  45. 

FIRST    PROCESS  SECOND    PROCESS 

a^4-4x  =  45  ar*  +  4aj  =  45 

/c2  +  4a;-45=0  a^  +  4x  +  4  =  49 

(a;-5)(a;  +  9)=    0  (a;  +  2)(a;  +  2)=7  •  7  or -7- -7 

.-.  a;-5=    0  .-.    a;  +  2  =  7  or  -7 

or                       a;  +  9=0  a;  =  7-2or-7-2 

.*.  a;  =  5  or  —   9  /.  a;  =  5  or  —  9 


FACTORING  113 

Explanation. —The  explanation  given  for  example  12  will  serve  for  the 
first  process. 

In  the  second  process,  it  is  seen  that,  by  adding  4  to  each  member  of  the 
equation,  the  first  member  will  become  the  square  of  the  binomial  (x  +  2). 
Solving  for  (a:  +  2)  as  for  x  in  previous  examples,  x  +  2—±l;  whence 
a;  =  db7-2  =  6or  -9. 

Sdggestion.  —  In  the  following  examples,  when  the  coefficient  of  the  first 
power  of  the  unknown  number  is  even,  either  of  the  above  processes  may  be 
used  ;  but  when  it  is  odd,  the  first  process  is  simpler. 

Solve  the  following  equations : 

43.  a.-2-6a;  =  40.  62.  ar^  +  4a;  +  3  =  0. 

44.  a^- 8  a;  =  48.  63.  ar^  +  6ar  +  8  =  0. 

45.  .x-2-5a;  =  -4.  64.  x2-9x  +  20-0. 

46.  x''-lx  =  l^.  65.  ic2  +  lla;  +  30  =  0. 

47.  x^ -}- 10  a;  =  56.  66.  a;^  +  a;  -  132  =  0. 

48.  a;'-' +  12 a;  =  28.  67.  Z2  =  4.x  +  x'. 

49.  a^~3ai  =  40.  68.  3a:  =  88 -ar^. 

50.  ar'-9a;  =  36.  69.  160  =  .x-^  -  6  a:. 

51.  a:^ _|_  11  a;  =  26.  70.  Ay  =  y--1^2. 

52.  x^  -  12  x  =  4.5.  71.  600  =  ^2  _  10,^^ 

53.  y^-Wy  =  5i.  72.  ^+16x-36  =  0. 

54.  y^-21y  =  A6.  73.  a;^  +  15a;  -  34  =  0. 

55.  a:2_i0a;  =  96.  74.  ?/2  -  8  ?/ -  84  =  0. 

56.  f-20y  =  96.  75.  y^ -2ay  +  a^  =  0. 

57.  2/2  +  12y  =  85.  76.  a;2  + 2^  +  6^=0. 

58.  2/^ +  42  =  13 y.  77.  x^  +  Aax +  Aa^=  0. 

59.  2/-  +  63  =  16t/.  78.  2='  +  22z  + 121=0. 

60.  v^-60  =  11  V.  79.  X-  -(a  +  h)x  +  a&  =  0. 

61.  y^  +  140  =  72 y.  80.  x^  +{c-^  d)x  +  ccZ  =  0. 


114  FACTORING 

81.  V+(a  +  2)a;  +  2a  =  0.  83.    x" -{a  -  d)x  -  ad  =  Q. 

82.  /-(c-n)?/  — nc  =  0.  84.    a?  —{h -\-'!)x +  1  h  =  0. 

85.  (2a;  +  3)(2a;-5)-(3a;-l)(a;-2)=l.      ' 

86.  (2a;-6)(3a;-2)-(5a;-9)(x-2)=4. 

87.  Solve  6x2  + 5a; -21=0. 

Solution 
6  x2  +  6  a;  -  21  =  0. 
Factoring,  §  131,  (2  a;  -  3)  (3  a;  +  7)  =  0. 

.-.  2  a;  -  3  =  0, 
or  3  X  +  7  =  0. 

.•.  X  =  I    or  —  |. 

Solve  the  following  equations: 

88.  3ic2  +  2x-l  =  0.  93.  Ix'  +  Qx-l^a. 

89.  5x2_,_4a;_i  =0.  94.  2 1)2- 9v- 35  =  0. 

90.  3!/'' +  2/ -10  =  0.  95.  6/ -222/ +  20  =  0. 

91.  3?/2_4y-4  =  0.  96.  30^*  + 13a; -30  =  0. 

92.  41/2  +  92/ -9  =  0.  97.  4a;2^-L3a,_i2  =  o. 

98.  Solve  the  equation  a?  —  2oi?  —  6x-^Q  =  0. 

Solution 

a;3_  2x2-5x4-6  =  0. 
Factoring  by  the  factor  theorem,  (x  —  l)(x  —  3)(x  +  2)  =  0. 
X  -  1  =  0,  or  X  -  3  =  0,  or  X  +  2  =  0. 
.•.  X  =  1,  or  3,  or  —  2. 

Solve  the  following  equations : 
99.   ar^  -  15  a.-^  +  71  a;  -  105  =  0.  101.    a;' -  12  a;  +  16  =  0. 

100.    a;3  _^  10 a;2  +  11  a;  -    70  =  0.  102.    a;^  -  19a;  -  30  =  0. 

lOS.    a;*  +  a.-^  -  21  a;2  _  a.  _,_  20  =  0. 

104.  a;*-7a;3  +  a;*  +  63a;-90  =  0. 

105.  a;*+8a;3-a;2_(58a;  +  60  =  0. 

106.  ar>-lla;*  +  45a;3-85a,-2  +  74a;-24  =  0. 


HIGHEST   COMMON   DIVISOR 


142.  1.  Name  all  the  numbers  that  will  exactly  divide  both 
a*  and  a?.     Which  of  these  is  of  the  highest  degree  ? 

2.  What  is  the  highest  divisor  common  to  6^  and  h^?  to  a? 
and  X? 

3.  Since  the  highest  divisor  common  to  a*  and  a?  is  o?,  to  h^ 
and  W  is  h^,  and  to  x^  and  x  is  x,  what  is  the  highest  divisor  com- 
mon to  a*&V  and  a?h^x  ? 

4.  What  is  the  highest  common  divisor  of  36a^&  and  90  a6^? 
What  prime  factors,  or  divisors,  are  common  to  36  a^h  and  90  ab^  ? 
How  may  the  highest  common  divisor  of  36a^6  and  90  a6^  be 
found  from  their  factors  ? 

143.  A  number  that  exactly  divides  each  of  two  or  more  alge- 
braic expressions  is  called  a  Common  Divisor  of  them. 

The  common  divisors  of  12  a^  and  4  a^  are  2,  a,  4,  a^,  2  a,  4  a,  2  a^, 
and  ia^. 

An  exact  divisor  of  an  expression  is  a  factor  of  it. 

Two  expressions  whose  only  common  divisor,  or  factor,  is  1  are 
said  to  be  prime  to  each  other. 

3  X  and  2  a  are  prime  to  each'other  ;  also  x  +  y  and  x  —  y. 

144.  That  common  divisor,  or  factor,  of  two  or  more  algebraic 
expressions  which  is  of  the  highest  degree  is  called  their  Highest 
Common  Divisor,  or  Highest  Common  Factor. 

The  highest  common  divisor,  or  factor,  of  12  a^  and  4  a^  jg  4  ^2, 
The  abbreviation  H.  C.  D.  is  used  for  Highest  Common  Divisor. 
The  highest  common  divisor  in  algebra  corresponds  to  the  greatest  com- 
mon divisor  in  arithmetic.  .  But  there  would  be  an  inaccuracy  in  applying 
the  term  greatest  common  divisor  to  literal  numbers,  since  letters  may  repre- 
sent any  numbers,  as,  for  instance,  fractions. 

116 


116  HIGHEST  COMMON  DIVISOR 

Thus,  if  ffl  =  I,  then  a?  =  \,  a?  =  \,  and  the  higher  the  degree  of  the  literal 
number,  the  less  will  be  its  arithmetical  value.  Consequently,  it  is  inaccurate 
to  speak  of  a^,  the  highest  common  divisor  of  a^  2  a*b,  and  3  a^,  as  their 
greatest  common  divisor,  because  a  may  represent  a  proper  fraction. 

145.  Pkinciple. —  The  highest  common  divisor,  or  factor,  of  two 
or  more  algebraic  expressions  is  the  product  of  all  their  common 
prime  factors. 

146.  To  find  the  highest  common  divisor  of  expressions  that  may 
be  factored  readily  by  inspection. 

Examples 
1.   What  is  the  highest  common  divisor  of  12  a^h^c  and  8  a^W(?  ? 

FIRST    PROCESS  SECOND    PROCESS 

12  a*b''c  =  3  X  2  X  2  X  aaaa  x  bb  x  c        12  a*b'c  =  4  a^b'^c  x  3  a" 
8a'6V  =  2  X  2  X  2  X  aa  X  666  X  ccc  8a^b^(f  =  ^a'Wc  x2b<? 


H.  C.  D.  =  2  X  2  X  aa  X  66  X  c  ^^a'bh     H.  C.  D.  =  4 d'b^ 

Explanation.  —  Since  the  highest  common  divisor  of  the  expressions  is 
the  product  of  all  their  common  prime  factors  (Prin.),  and  since  the  only 
prime  factors  common  to  the  given  expressions  are  2,  2,  a,  a,  b,  6,  and  c, 
their  product,  'ia^b^c,  is  the  highest  common  divisor. 

Suggestion. — Frequently  the  work  may  be  abridged  by  grouping  com- 
mon factors,  as  in  the  second  process.  Since  3  a^  and  2  bc^  are  prime  to 
each  other,  4  a^b'^c  must  contain  all  the  common  factors,  and  be  the  highest 
common  divisor. 

2.   What  is  the  H.  C.  D.  of  3x^-3xy^  and  a^  -  2  ar'y  +  a?j^  ?    ' 

PROCESS 

3a:^  —  3xy^  =3x(x  +  y)(x  —  y) 

a?  —2  3?y  4-  xy^  =    x(x  —  y)(x  —  y) 

.-.  H.C.D.=    x(x-y) 

Explanation. — For  convenience  in  selecting  the  common  divisors,  the 
expressions  are  resolved  into  their  simplest  factors. 

Since  the  only  common  prime  factors  are  x  and  (x  —  y),  the  highest 
common  divisor  sought  is  their  product,  x{x  —  y)  (Prin.). 


HIGHEST  COMMON  DIVISOR  117 

Rule.  —  Separate  the  expressions  into  their  p>rime  factors. 

The  product  of  all  the  common  prime  factors,  each  factor  being 
taken  the  least  number  of  times  it  occurs  in  any  of  the  given  expres- 
sions, is  the  highest  common  factor. 

The  factors  that  enter  into  the  H.  C.  D.  can  often  be  selected  without 
actually  separating  the  expressions  into  their  prime  factors. 

3.   What  is  the  H.  C.  D.  of  5  aV  -  5  6V  and  a^a?  -  Wv?  +  aV 

-6y? 

Solution 

5  a2c2  _  5  h2c2  =  5  c2(a2  _  62) 

d^x^  -  &%s  +  aV  -  foy  =  (x«  +  f)  {d^  -  b') 
.-.  H.  C.  D.  =  a2  -  62 

Find  the  highest  common  divisor  of 

4.  10  2(?y^,  10  a-y,  and  15  xy*z. 

5.  l()a%\  21  a*b*,  and  35  a*b\ 

6.  Sm^n^,  28  m%*,  and  56m*w^, 

7.  4:b^cd,  6  6V,  and  24  aftc^. 

8.  10  (x  —  2/) V  and  15  (z  —  y)(x  —  y)^. 

9.  4(a  +  bf{a  -  h)  and  b{a+  h)\a  -  6)2. 

10.  3  (a^  -  62)2  and  a  (a  -  6)  (a^  -  6^). 

11.  x-2  -  2  a;  -  15  and  a^  -  a;  -  20. 

12.  x*  —  y*,  a?  —  }f,  and  x-\-y. 

13.  a^  +  7  a  +  12  and  a'^  +  5  a  +  6. 

14.  a^  +  2/^  and  x^  +  2xy  -\-  y^. 

15 .  a^  —  a?  and  a^  —  2  aa;  +  a:^. 

16.  a2-62  and  a2  +  2a6  +  62. 

17.  x*  +  a:2y2  ^  yi  a^jj(j  a^  ^_  a-y  _^  ^2. 

18.  a^  +  2/^,  a^  +  i/',  and  a^?/  +  xy^- 

19.  a^  +  a26*  +  6»  and  3  a^  -  3  a62  +  3  6*. 

20.  a^  —  a?,  a^-\-2ax  +  x',  and  a^  +  a^. 

21 .  ax  —  y  +  xy  —  a  and  aa;'^  +  a^y  —  a  —  y. 

22.  a^6  —  6  —  a'^c  +  c  and  a6  —  ac  —  6  +  c. 


118  HIGHEST  COMMON  DIVISOR 

23.  1-4:X^,  l  +  2x,  and  A  a -16  ax'. 

24.  (a  -  b)  (h  -  c)  and  (c  -  a)  {a?  -  6-). 

25 .  24  af*/  +  8  x'f  and  8  a^i/^  _  g  r^^^ 

26.  6  if2  +  a;  -  2  and  2  a^  -  11  a;  +  5. 

27.  16a:2_25  and  20ar'-9a;-20. 

28.  a^  -  6  .T  +  5  and  a^-5a^  +  7a;-3. 

29.  a? -A  and  a^  -  lOa^  +  31a;- 30. 

30.  x'-^  and  a;«  -  12  a;^  ^  41 3.  _  42, 

31.  a;3  _  4  a;  +  3  and  ar'  +  ar  —  37  a;  +  35. 

147.  To  find  the  highest  common  divisor  of  expressions  that  can- 
not be  factored  readily  by  inspection. 

1.  What  are  the  exact  divisors  of  a&?  Will  they  be  factors 
of  2  times  a&  ?  of  a  times  cib  ?  of  m  times  ah  ? 

2.  If  a  number  is  an  exact  divisor  of  an  expression,  what  will 
be  its  relation  to  any  number  of  times  the  expression  ? 

3.  What  common  divisor  have  ctiX  and  ay  ?  any  number  of  times 
ax  and  ay,  as  tn  •  ax  and  n  •  ay  ? 

4.  If  two  numbers  have  a  common  divisor,  what  divisor  has 
their  sum  ?  their  difference  ?  the  sum  or  difference  of  any  n\im- 
ber  of  times  the  numbers  ? 

5.  What  is  the  highest  common  divisor  of  2am{x+y')  and 
Shm{x  +  y)?  How  will  it  be  affected,  if  the  second  number  is 
multiplied  by  7  or  2:  ?  by  2  or  a  ?  How  will  it  be  affected,  if  the 
first  number  is  multiplied  by  5  ?  hy  b? 

6.  By  what  numbers  may  one  of  two  expressions  be  multi- 
plied without  affecting  their  highest  common  divisor  ? 

7.  How  will  the  highest  common  divisor  of  2am(a;  +  ^)  and 
3bm{x  +  y)  be  affected,  if  the  first  number  is  divided  by  2  ? 
by  a  ?  by  m  ?  by  (a;  +  y)?  How,  if  the  second  number  is  divided 
by  6  ?  by  m  ? 

8.  By  what  numbers  may  one  of  two  expressions  be  divided 
without  affecting  their  highest  common  divisor  ? 


HIGHEST  COMMON  DIVISOR  119 

148.  Principles.  —  1.  A  divisor  of  an  expression  is  a  divisor 
of  any  number  of  times  the  expression.     Hence,  by  §  85, 

2.  A  common  divisor  of  two  expressions  is  a  divisor  of  their  sum, 
of  their  difference,  and  of  the  sum  or  the  difference  of  any  number 
of  times  the  expressions  ;  also, 

3.  The  highest  common  divisor  of  two  expressions  is  not  affected 
by  mxdtiplying  or  dividing  either  of  them  by  numbers  that  are  not 
factors  of  the  other. 

Examples 
1.    Find  the  H.  C.  D.  of  x'  +  bx  +  Q  and  4a^  +  21  x^  +  SOa; -f  8. 

PROCESS 

ar^  +  5  a;  +  6)4  ar^  +  21  x2  4-  30  a;  +  8(4  »  +  1 
4  a^  +  20  ar*  +  24  a; 

x^+    6x  +  8 
a?+    5a;  +  6 

X  +  2)v?  +  5  a;  +  6(a;  +  3 
x^  +  2x 

3x  +  6 
.-.  H.C.D.  =  a;  +  2.  3  x- +  6 

Explanation.  —  Since  the  highest  common  divisor  cannot  be  higher  than 
05^  +  5  X  +  6,  it  will  be  a;2  ^  5  a;  ^  6,  if  x^  +  5  x  +  6  is  exactly  contained  in 
4  x^  +  21  x2  +  30  X  +  8.  By  trial,  it  is  found  that  it  is  not  exactly  contained 
in  4  X*  +  21  x^  -I-  .30  x  +  8,  since  there  is  a  remainder  of  x  +  2.  Therefore, 
x^  +  5  X  +  6  is  not  the  highest  common  divisor. 

Since  x'^  +  5  x  +  6  contains  the  highest  common  divisor,  (4  x  +  1)  times 
x2  +  5x  +  6  will  also  contain  the  highest  common  divisor  (Prin.  1);  and 
since  both  4  x"  +  21  x^  +  30  x  +  8  and  (4  x  +  1)  (x^  +  5  x  +  6)  contain  the 
highest  common  divisor,  their  difference,  x  +  2,  must  contain  the  highest 
common  divisor  (Prin.  2).  Hence,  the  highest  common  divisor  cannot  be 
higher  than  x  +  2. 

X  +  2  will  be  the  highest  common  divisor,  if  it  is  exactly  contained  in 
x2  +  6x  +  6,  since,  if  it  is  contained  in  x2  +  5x  +  6,  it  will  be  contained 
in  any  number  of  times  x^  +  5  x  +  6,  as  (4  x  +  1)  (x^  -f  5  x  +  6)  (Prin.  1); 
and  in  the  sum  of  (4  x  +  l)(x2  +  5  x  +  6)  and  x  +  2,  or  4  x^  +  21  x^  +  30  x  +  8 
(Prin.  2).  By  trial,  x  +  2  is  found  to  be  exactly  contained  in  x^  +  5  x  +  6. 
Therefore,  x  +  2  is  the  highest  common  divisor  of  the  given  expressions. 


120  HIGHEST  COMMON  DIVISOR 

2.    Find  the  H.  C.  D.  of  Oa^  +  33a;-63  and  2a;3_,_llar'-a;-30. 


PKOCESS 


3)6  a?  +  33  X  -  63 
2  a^  +  11  a;  -  21 


2cc3  +  llic2_       a._  3Q(a. 
2  x^ -f  11  x^  _  21  a; 


10)20  a;  -  30 

2  a;-    3)2  a;2  ^  11  a;  -  21(a;  +  7 
2ar'-    3a; 


H.  C.  D.  =  2  a;  -  3. 


14  a;  -  21 
14  a;  -  21 


Suggestion. — Since  only  common  factors  are  sought,  factors  that  ar« 
not  common  to  the  given  expressions,  as  3  p.nd  10,  may  be  rejected  from  any 
expression  before  it  is  used  as  a  divisor  (Prin.  3). 

3.  Find  the  H.  C.  D.  of 

2  a;3  _^  5  ar^  _  22  a;  +  15  and  ba? +  l^v? -dZx  +  lQ. 


PROCESS 

2a;3_^5aJ!_22a;  + 15)5  a.-^  +  18  a;''  - 

9 


33  a;  + 10 


10a;3_^36a^_  66  a; +  20(5 
10  a:^  +  25  a;^  -  110  g;  +  75 
11)11  a;^+  44  a; -55 
01?  ^r      4a;—  5 

ar^  +  4  a;  -  5)2  a;3  ^  5  3,2  _  22  a;  +  15(2  a;  -  3 
2  a;3  +  8  a;2  -  10  a; 


H.  C.  D.  =  a;2  _^  4  a- _  5. 


-  3  a;2  -  12  a;  +  15 

-  3  a;^- 12  a; +  15 


Suggestion.  —  When  the  first  term  of  the  divisor  is  not  contained  in  the 
first  term  of  the  dividend  an  integral  number  of  times,  fractional  quotients 
may  be  avoided  by  nmltiplying  the  polynomial  taken  for  the  dividend  by 
some  number  not  a  factor  of  the  divisor  (Prin.  3).  In  the  above  example 
the  simplest  factor  that  may  thus  be  introduced  is  2,  if  5  x^+lS  a:'-  — 33x  +  10 
is  taken  for  the  dividend  ;  or  5,  if  2  x^  +  5  a;^  _  22  x  +  15  is  taken  for  the 
dividend. 


HIGHEST   COMMON  DIVISOR  121 

4.  Find  the  H.  C.  D.  of 

30  amx^  —  21  amx  —  99  am  and  42  ahoi?  +  33  ab:i?  —  45  ahx, 

PROCESS 

30  am2(?  —  21  amo;  -  99  am )  42  aha?  +  33  a&a;^  —  45  ahx 

Eeject  m  (Prin.  3).  Eeject  hx  (Prin.  3). 

30 aar^  -  21  ox  -  99 a)42  ax"  +  33  aa;  -  45  a 

Reserve  the  common  factor  3  a  as  a  factor  of  the  H.  G.  D. 

lOar'-    Ix-    ^Z)Ux'-\-llx-    15 
_7 5 

70ar'-49a;-231)70a,-2  +  55a;-   75(1 
70  ar^- 49  a; -231 

52)  104  a; +  156 

2a;+      3 
2a; +  3)10x2 -7a;-   33(5a;_ll 
.-.  H.C.D.  =  3a(2a;  +  3).  10a;2-7a;-    33 

ScGGESTioN.  —  Since  each  of  the  polynomials  contains  a  factor  not  found 
in  the  other,  these  two  factors  may  be  rejected  (Prin.  3).  Consequently,  m 
is  rejected  from  the  first  polynomial,  and  bx  from  the  second. 

To  simplify  the  process  the  common  factor  3  a  is  removed  and  reserved  as 
a  factor  of  the  H.  C.  D. 

5.  Find  the  H.C.D.  of  9a;2-35a;  +  24  and  29»-8a;«-15. 

PROCESS 

-  8  a;2  ^  29  a;  -  15 
9a;2-35a;  +  24 

a?-    6a;+    9)9a;2  -  35«  +  24(9 

9a;2_54a;  +  81 

19)  19  a; -57 

X-   3)ar'-6a;  +  9(a;-3 
.-.  H.  C.  D.  =  a;  -  3.  a;'-6a;  +  9 

ScGGBSTioN.  —  Since  the  sum  or  the  difference  of  two  expressions  con- 
tains their  highest  common  divisor  (Prin.  2),  it  is  evident  that  at  the  outset 
a  simpler  expression  that  will  contain  the  highest  common  divisor  may  be 
obtained  by  adding  the  given  expressions,  giving  oj'^  —  6  x  +  9. 


122  HIGHEST  COMMON  DIVISOR 

Find  the  H.  C.  D.  of 

6.  a^  +  2  a;  -  24  and  2  a^  +  7  X  -  30. 

7.  2  x^  _  a.  _  21  and  4x2  + 4  a; -63. 

8.  3x2+.10x-8  and  6x2-7x  +  2. 

9.  2x3-6x2  +  7x-6  and  2x3  +  4ar'-3x  +  9. 
10.   x3^9aJJ4.26x  +  24  and  2x3_^14a^_^20x. 

11.   Find  the  H.  C.  D.  of 
3  aa?  —  4  ax^  —  13  ax  +  14  a  and   3  ah:x?  +  5  ab^  —  10  abx  —  42  ah. 

FIRST    PROCESS 

3  ox^  —  4  aa^  -  13  ax  +  14  a)3  aha?  +  5  ahx^  —  10  a6x  —  42  ah 

Reserve  the  common  factor  a  as  a  factor  of  the  H.  C.  D. 

3a^_4ar=_13a;+14)36x3+56x2-10  6x-42  6(6 
36x3-46x2- 13  &X4-146 
6)96x2+   'Sbx-bQb 
9x2  _j.  3^  _56)3a^_  4x2-13x+14 

3 

9.r'-12x2-39x+42(x 
9x^+  3x2-56x 

-15x2+17x+42 
-15x2+17x+42)9x2+  3x-  56 

_5 

46x2+15x-280(-3 
45x2-51  x-126 
22)66  X- 154 

3x-     7)-15x2+17x+42(-5x-6 

-15x2+35x 

-18X+42 
.-.  H.C.D.  =a(3x-7).  -18x+42 

Since  the  arrangement  of  the  dividend,  divisor,  and  quotient 
may  be  either:  Divisor)  Dividend  (Quotient;  or  Quotient)  Divi- 
dend ( Divisor ;  by  using  these  two  arrangements  alternately,  the 
above  process  may  be  more  compactly  written  as  follows : 


HIGHEST  COMMON  DIVISOR 


123 


SECOND    PROCESS 


3  aoi?  —  4  (vx-  —  13  ax  +  14  a)3  ahx^  +  5  ahy?  —  10  a&ic  —  42  ah. 
Reserve  the  common  factor  a  as  a  factor  of  the  H.C.D. 


3ar5_    4  a,-^  -  13  a;  +  14 
3 

9  a^  -  12  ar^  -  39  a;  +  42 
9ar'+    3a^-oGa;' 

3bs(^  +  5ba^-  10  &x  -  42  & 
3  6ar*  -  4  ftar^  -  13  6a;  +  14  & 

h)9ba^+    3te-56& 

X 

9a^-i-    3x    -56 
5 

-  15  a;-'  +  17  a;  +  42 

-  15  a^  +  35  X 

45ar'  +  15a;    -280 
45a;2-51a;    -126 

22)66  a;    -154 

—  5x 

3a;    -7 
.-.  H.C.D.  =  a  (3  a; -7). 

-6 

-  18  a;  +  42 

-  18  a;  +  42 

-3 


Suggestion.  —  When  the  quotient  consists  of  more  than  one  term,  for 
convenience  each  term  is  placed  opposite  the  corresponding  part  of  the  divi- 
dend or  product. 

Rule. — Divide  one  expression  by  the  other,  and  if  there  is  a 
remainder,  divide  the  divisor  by  it;  then  divide  the  preceding  divisor 
by  the  last  remainder,  and  so  on,  until  there  is  no  remainder.  The 
last  divisor  will  be  the  highest  common  divisor. 

If  any  remainder  does  not  contain  the  letter  of  arrangement,  the 
expressions  have  no  common  divisor  in  that  letter. 

If  more  than  two  expressions  are  given,  find  the  highest  common 
divisor  of  any  two,  then  of  this  divisor  and  another,  and  so  on.  Tlie 
last  divisor  will  be  the  highest  common  divisor. 

1.  If  either  expression  contains  a  monomial  factor  not  found  in  the  other, 
it  should  be  rejected  before  beginning  the  process. 

2.  A  common  factor  of  the  expressions  should  be  removed  before  begin- 
ning the  division,  but  it  must  appear  as  a  factor  of  the  highest  common  divisor. 

3.  When  necessary,  to  avoid  fractional  quotients,  any  dividend  or  divisor 
may  be  multiplied  or  divided  by  any  number  not  a  factor  of  the  other. 

4.  The  highest  common  divisor  has  an  ambiguous  sign.  For,  if  a  positive 
divLsor  is  contained  in  a  dividend,  the  same  negative  divisor  also  will  be  con- 
tained in  that  dividend,  but  the  signs  of  the  quotient  will  be  changed.  It  is 
not  customary  to  write  both  divisors. 


124  HIGHEST  COMMON  DIVISOR 

149.  The  principle,  that  the  exact  divisor  reached  by  the  process 
given  in  the  rule  is  the  highest  common  divisor,  may  be  proved  as 
follows : 

Let  A  and  B  represent  any  two  polynomials  freed  of  monomial  factors, 
the  degree  of  B  being  not  higher  than  that  of  A. 

Divide  A  by  B,  and  let  the  quotient  be  m  and  the  remainder  D  ;  divide  B 
by  D,  and  let  the  quotient  be  n  and  the  remainder  E ;  divide  D  by  E,  and 
let  the  quotient  be  r  and  the  remainder  zero ;  that  is,  let  E  be  an  exact 
divisor  of  D. 

It  is  to  be  proved  that  E  is  the  highest  common  divisor  of  A  and  B. 

PROCESS 

B)A{m 
mB 
D)B(n 
nP 
E)D{r 
rE 
0 

Since  the  minuend  is  equal  to  the  subtrahend  plus  the  remainder, 
A  =  mB  +  D,  and  A  -  mB  =  D ; 
B=  nD  +  E,  and  B-  nD  =  E;  and  D  =  rE. 

Since  the  division  has  terminated,  E  is  a,  common  divisor  of  D  and  nD 
(Prin.  1)  ;  also  of  D  and  nD+E,  or  B  (Prin.  2)  ;  also  of  B  and  mB  (Prin.  1) ; 
also  of  B  and  mB  +  D,  or  A  (Prin.  2)!  That  is,  JE"  is  a  common  divisor  of  B 
and  A. 

Every  common  divisor  of  A  and  B  is  a  divisor  of  mB  (Prin.  1)  ;  and  of 
A  —  mB,  or  D  (Prin.  2),  Therefore,  every  common  divisor  of  A  and  B  is  a 
divisor  of  nD  (Prin.  1)  ;  and  of  B  —  nD,  or  E  (Prin.  2). 

But,  since  no  divisor  of  E  can  be  of  higher  degree  than  E  itself,  E  is  the 
highest  common  divisor  of  A  and  B.  ' 

150.  The  principle,  that  the  highest  common  divisor  of  several 
expressions  may  be  obtained  by  finding  the  highest  common 
divisor  of  two  of  them,  then  of  this  result  and  a  third  expression, 
and  so  on,  may  be  proved  as  follows : 

Let  P  be  the  highest  common  divisor  of  A  and  B,  and  Q  the  highest  com- 
mon divisor  of  P  and  a  third  expression  C. 

Then,  since  P  contains  all  the  common  factors  of  A  and  B,  and  Q  con- 
tains of  these  particular  factors  only  such  as  are  factors  of  C  also,  Q  is  the 
highest  common  divisor  of  A,  B,  and  C. 

This  method  may  be  extended  to  embrace  any  number  of  expressions. 


HIGHEST  COMMON  DIVISOR 
Find  the  H.  C.  D.  of 


125 


12.  2a:3-7aj2  +  2a;  +  3  and  2.'K3  +  7a^-5a;-4. 

13.  9a^  +  18a:2-a;-10  and  3a^  +  13a;2  +  2a;-8. 

14.  l-2a;-5a^  +  6a^  and  l  +  5a;  +  2if2-8af'. 

15.  l-4a;  +  a^  +  6a^  and  l  +  3a;-6a^-8a^. 

16.  \-x-Ux'-{-24.v?  and  d&a?-2^a?  +  x  +  l. 

17.  m^  —  4 nv?  —  20 m  +  48  and  m^  —  m^  —  14 m  +  24. 

18.  3a''  +  20a2-a-2  and  3a»+17a2  +  21a-9. 

19.  8ax2  +  22aa;  +  15a  and  66a^  +  ll  6a;  +  3&. 

20.  20  ft^c  -  2  &c  -  4  c  and  8  a^ft^c  -  4  a?bc  +  a^c. 

21.  21  a.r  — 17  aa!^  —  5  aa^  +  aa;*  and  7  aa;  +  34  aa^  —  5  aa^. 

22.  a^-7a;  +  6,  a-*  -  2aj3  -  9a^  +  18a;,  a;3  + a;^ -4a;-4. 

23.  x^-5x  +  4:,  x*-2x'^  +  l,  a;«  +  4a^-3a;-2. 

24.  l+4a^  +  5a^,  2  +  5a;  +  3x*,  a;« - 4 a;*  +  5 ar^ - 2. 

25.  3  +  ar-8a^  +  4a^,  3 -8a;-8a^4-8a^,  16 a;* - 48 a^  +  81. 

26.  c^-6a^-5x-U,  x3-10a:2^20a:  +  7,  a;* - 310 a; - 231. 

27.  FindtheH.C.D.  of  a;«+a^+a^-a;-2  and  2a^+a;^-a^-a:*-l. 


PBOCESS    BY    DETACHED   COEFFICIENTS 


1+0+1+1+0-1-2 

1-2-3-1+2+3 

2+4+2-2-4-2 

2-4-6-2+4+6 

8)8  +  8  +  0-8-8 

1+1+0-1-1 

.-.  H.C.D.  =  a^+a^-a;-l. 


2+1+0-1-1+0 

2+0+2+2+0-2 


1-2-3-1+2+3 


1+1+0-1-1 
-3-3+0+3+3 
-3-3+0+3+3 


Find  the  H.  C.  D.  of 

28.  ar*-a^-2a:3_a^  +  a,  +  2  and  a^  +  3a^  +  3a^  +  a^-a;-l. 

29.  af +  a;*-a^-7a;-4  and  2ar^  +  3a;*  +  3ar' +  3ar' -  7a;-  4. 


126  LOWEST  COMMON  MULTIPLE 

30.  a^-2x*-2x''-llx'-x-15sind2aT'-7x*+A2^-15x^+x-10. 

31.  a'-3a*-3a^-3a'-19a-15sLnd  a'+3a*-3a^+9a^-a-15, 

32.  5  a^-\-a*- 11  a^ +9  a^-Sa+4:  and  2  a^-a^-iT  a^-f  8  a--4  a. 

33.  x^  —  5x  +  4:  and  ic*  —  a:'  —  3 a^  —  5 a;  — 12. 

34.  a'  +  3  a^  -  2  a  -  6  and  a^  +  4  a*  +  4  a''  +  4  a^  -  a  -  12. 

35.  1  -  4 a^  +  3a*  and  1  +  a  -  a'^  -  5a^  +  Aa*. 

36.  2  —  a  +  3a^  +  5a^-a*  and  4  —  4 a  +  a^  —  9 a*. 

37.  f-\-13y^  +  20y-U  and  7  -  3?/ -  20?/=' +  2?/- y*. 

38.  6x3-lla^-35a;,  30  ar^  -  115  x  +  35,  23^  -  dx"  -  ox-7. 


oV»ic 


LOWEST   COMMON   MULTIPLE 


151.  1.  What  number  exactly  contains  2,  5,  a,  and  b,  or  is  a 
multiple  of  2,  5,  a,  and  b  ? 

2.  What  different  prime  factors  must  enter  into  every  number 
that  will  contain  4  a^b,  a^b^,  and  10  ab^,  or  must  be  found  in 
every  common  multiple  of  Aa^b,  a^b'^,  and  10  a6^? 

3.  What  is  the  lowest  power  of  a  that  common  multiples 
of  4  a^b,  d^b^,  and  10  a¥  can  contain  ?  What  is  the  lowest  power 
of  fe  ?    of  2  ?   of  5  ? 

What,  then,  is  the  lowest  common  multiple  of  4  a?b,  a^b',  and 
10  aW? 

To  what  is  the  lowest  common  multiple  of  two  or  more  expres- 
sions equal  ? 

152.  An  expression  that  exactly  contains  each  of  two  or  more 
given  expressions  is  called  a  Common  Multiple  of  them. 

6  abx  is  a  common  multiple  of  a,  3  6,  2  x,  and  6  abx.  These  numbers 
may  have  other  common  multiples,  as  12  abx.,  6  a^6%,  18  a*6x'^,  etc. 


LOWEST  COMMON  MULTIPLE  127 

153.  The  expression  of  knvest  degree  that  will  exactly  contain 
each  of  two  or  more  given  expressions  is  called  their  Lowest 
Common  Multiple. 

6  abx  is  the  lowest  common  multiple  of  a,  3  ft,  2  a;,  and  6  abx. 

The  abbreviation  L.  C.  M.  is  used  for  Lowest  Common  Multiple. 

The  lowest  common  multiple  in  algebra  corresponds  to  the  least  common 
multiple  in  arithmetic.  But,  since  letters  may  represent  any  numbers,  as, 
for  instance,  numbers  not  prime  to  each  other  or  fractions,  the  term  least 
is  not  applicable  to  algebraic  common  multiples. 

Thus,  the  algebraic  lowest  common  multiple  of  a^b'^,  ab^,  and  bx  is  a^b'x. 
If  a  =  4,  6  =  3,  and  x  =  2,  a'^b^x,  the  lowest  common  multiple  of  the  given 
expressions,  is  equal  to  864.  If,  however,  the  values  of  a,  ft,  and  x  are  sub- 
stituted for  those  letters,  the  given  expressions  become  144,  108,  and  6  ;  and 
their  least  common  multiple  is  432. 

It  is  thus  seen  that  tlie  lowest  common  multiple  of  two  or  more  expressions 
is  not  necessarily  their  least  common  multiple. 

154.  Principle.  —  The  lowest  common  multiple  of  two  or  more 
algebraic  expressions  is  the  product  of  all  their  different  prime  fac- 
tors, using  each  factor  the  greatest  number  of  times  it  occurs  in  any 
of  the  expressions. 

155.  To  find  the  lowest  common  multiple  of  expressions  that  may 
be  factored  readily  by  inspection. 

Examples 
1.   What  is  the  L.  C.  M.  of  12  x'yz*,  6  a^xf,  and  8  axyz^  ? 

PROCESS 

12x^yz'   =2'2'3-x^-y-^ 
6  a^xy'^  =2  •  3  •  a^  •  X  •  y'^ 
8  axyz^  =2  -2  '2  ^  a  •  x  -  y  •  '^ 


L.  C.  M.  =  2  .  2  .  2  .  3  .  a=^  •  ar'  •  /  •  «*  =  24  aV^z* 

Explanation.  — The  lowest  common  multiple  of  the  numerical  coefficients 
is  found  a.s  in  arithmetic.     It  is  24. 

The  literal  factors  of  the  lowest  common  multiple  are  each  letter  with  the 
highest  exponent  it  has  in  any  of  the  given  expressions  (Prin.).  They  are, 
therefore,  a'^,  x^,  j/2^  and  2*. 

The  product  of  the  numerical  and  literal  factors,  24  a'^x^y'^z*,  is  the  lowest 
common  multiple  of  the  given  expressions. 


128  LOWEST  COMMON  MULTIPLE 

2.   What  is  the  L.  C.  M.  of  or'  -  2 »?/  +  /,  x?  -  y^,  and  a^  -f  2/»? 

PROCESS 

x^  -  2xy  +  y^  =  (x  -  y)(x  -  y) 
a^-y^  ={x-y)  {x  +  y) 
a^  +  f  —  {x  +  y){x^  —  xy  +  f) 

L.  C.  M.  =(x-y)\x+y){x'-xy,+  f) 

=  (a;  -  y)\:»?  +  f) 

Rule.  —  Factor  the  expressions  as  far  as  may  he  necessaTy  to 
discover  their  different  prime  factors. 

Find  the  product  of  all  their  different  prime  factors,  using  each 
factor  the  greatest  number  of  times  it  occurs  in  any  of  the  given 
expressions. 

The  factors  of  the  L.  C.  M.  may  often  be  selected  without  separating  the 
expressions  into  their  prime  factors. 

Find  the  L.  C.  M.  of 

3.  a^x^y,  a^xtf,  and  aa^y. 

4.  10  a^ft-V,  5  aft^c,  and  25  6Vd3. 

5.  16a'b\  24  d^de,  and  36  a^b^d^^. 

6.  IS  a'^br-,  12p^g^r,  and  b^ab^p^q. 

7.  a/*"*!/^,  ic^-y,  aj^'V,  and  x'^+^y. 

8.  a?  —  y^  and  a?  -\- 2  xy  -\- y^. 

9.  a?  —  y^  and  a?  —  2xy  +  y^. 

10.  X?  —  y^,  X?  -\-2  xy  +  y^,  and  .r^  —  2  a;?/  +  y^. 

11 .  a^  —  n^  and  3  a^  +  6  a^n  +  3  an^. 

12.  a;"-!  and  a2a^  +  a'-6^a;2-62 

13.  a^  +  1,  ab  —  b,a^  +  a,  and  a"^  —  1. 

14.  2x  -\-  y,2xy  —  y"^,  and  Ax^  —  y^. 

15.  1  +  a;,  a;  —  a^,  1  +  a^,  and  a^(l  —  x). 

16.  3  +  a,  9  —  a^,  3  —  a,  and  5  a  + 15. 

17.  a  —  b,h  —  c,b  +  a,  and  a^  —  b^. 


LOWEST  COMMON  MULTIPLE  129 

18.  2x  +  2,  5x  —  5,  Sx  —  3,  and  ccF  —  l. 

19.  3x-9y,  3a^  +  27r,  and  2x-\-6y. 

20.  16  62  _  1,  12  62  _|_  3  5,  20  6  -  5,  and  2  6. 

21.  1-23^  +  3;',  (l-ic)2,  and  l  +  2a;  +  a^. 

22.  1  -  a,  1  +  a,  1  +  a^,  1  +  a*,  and  1  +  al 

23.  xy  —  'tf,  a?  +  xy,  ocy-^y^,  and  a^  +  2/^. 

24.  a?  —  y^,  x'-\-xy-\-'if,  and  a?  —  xy. 

25.  6-  -  5  6  +  6,  6^  _  7  5  +  10,  and  6^  _  lo  6  +  16. 

26.  a^+ 7x  —  8,  ic^  _  1^  a;  ^aj2^  and  3cw^— 6ax  +  3a. 

27.  a^  —  x^,  a  —  2x,  a^  -\-2 ax,  and  a^  —  3  a^o;  +  2 aa^. 

28.  m^  —  a^,  y/i^  +  7n«,  m^  +  mx  +  aj^,  and  (m  +  3;)a^. 

29.  a^-3a;  +  2,  ar'  +  4a;  +  4,  a^  +  3a;  +  2,  and  ar*  — 1. 

30.  a;2  —  y^,  X*  +  a?y^  -\- y*,  a^  +  i/^,  and  x^  +  xy  +  y\ 

31.  a;^  +  a;2y  +  x'?/2  +  ?/3  and  oi?  —  x^y  +  xy^  —  y^. 

32.  a2^4a  +  4,  0^  —  4,  and  a*  — 16. 

33.  a^-{p-\-  cf,  62  -  (c  +  a)2,  and  (?-{a^  Vf. 

34.  1  —  a  +  a^,  1  +  a  +  a^,  and  1  +  a^  -f  a*. 

35.  a^  4-  4  and  a*  —  2  a^  +  4  a  —  4. 

36.  a«  -  6^  and  a*  +  a^62  +  6*. 

37.  a;^  +  y^  and  c^^  —  b'^y'^  +  a^?/^  —  62a;2. 

38.  a^-2  a?b  +  a262  -  9  6^  and  a*  +  5  a'b''  +  9  6*. 

39.  a*  —  a?  +  1,  a^  +  1,  a*  +  a^  +  1,  and  a*  —  1. 

40.  Find  the  lowest  common  multiple  of  aj^  +  6a^  +  5a;  — 12 

and  a:3_8a^  +  19a;-12. 

Ik 

Solution 
x8  +  6a;2+    5a;-12=(«-l)(x2  +  7a;+ 12) 

=  (x-l)(x  +  3)(«  +  4). 
a;8  _  8 x2  +  19 X  -  12  =  (x  -  l)(x2  -  7 X  +  12) 

=  (x-l)(x-3)(x-4). 
.-.  L.  C.  M.  =(x-l)(x  +  3)(x-3)(x  +  4)(x-4) 

=  (x-l)(x2-9)(x2-16). 

ALG.  — 9 


180  LOWEST  COMMON  MULTIPLE 

Suggestion.  — In  solving  the  following  the  Factor  Theorem  will  be  found 
useful. 

41.  a^-6x^  +  llas-6  2ind  x^-9x'  +  26x-24:. 

42.  ar*  -  5 x2  _  4  a;  +  20  aud  a^  -h  2ar^  -  25 a;  -  60. 

43.  cc^ -f-3a^  — 4  and  a^  + a^  — ic  — 1. 

44.  a^-4:a^  +  5x-2  Siud  a^-8x^  +21X-18. 

45.  a^  +  5a;2  +  7a;  +  3  and  a^ —  7 x^  —  5x  +  75. 

46.  a^  +  2a^-4a;-8,  a^-a^-8a;+ 12,  a^  +  4a^- 3a;- 18. 

47.  ar'-9a;2_^23a;-15,  a^  +  x'-lT x  +  15,  x^+yar^  +  jaj-m 

48.  a^  +  Tx'  +  Ux  +  S,  a.-3  +  3a^-6a;-8,  x^ +  x^ -lOx -^S. 

156.  To  find  the  lowest  common  multiple  of  expressions  that  can- 
not be  factored  readily  by  inspection. 

x'-Sx^   2  =  (x-l){x-2).  (1) 

a^_5a;+    6  =  (x-2)(x-3).  (2) 

a;2_7a;  +  12  =  (a;-3)(a;-4).  (3) 

L.  C.  M..  =  (x-  l){x  -2)(x-  3)(x  -  4). 

1.  Find  the  lowest  common  multiple  of  expressions  (1)  and  (2) 
from  their  factors ;  from  the  product  of  their  factors.  By  what 
factor  of  the  two  expressions  must  the  product  be  divided  to 
obtain  the  lowest  common  multiple? 

2.  How,  then,  may  the  lowest  common  multiple  of  two  expres- 
sions be  found  ? 

3.  Since  (x  —  l)(x  —  2)  (x  —  3)  is  the  lowest  common  multiple 
of  the  first  two  expressions,  what  factor  of  the  third  expression 
must  the  lowest  common  multiple  of  all  the  expressions  contain  ? 

157.  Principles.  —  1.  The  lowest  common  multiple  of  two  ex- 
pressions is  equal  to  their  product  divided  by  their  highest  common 
divisor;  or,  it  is  equal  to  either  of  them  multiplied  by  the  quotient 
of  the  other  divided  by  the  highest  common  divisor. 

2.  The  lowest  common  multiple  of  several  expressions  may  be 
obtained  by  finding  the  lowest  common  multiple  of  two  of  them  ;  then 
of  this  result  and  a  third  expression;  and  so  on. 


LOWEST  COMMON  MULTIPLE  131 

Proof  of  Principle  1. 

Let  F  be  the  highest  common  divisor,  or  factor,  of  A  and  B,  and  L  their 
lowest  common  multiple.  Let  F  be  contained  a  times  in  A  and  6  times  in  B, 
or  let  A  =  aF  and  let  B  =  bF. 

It  is  to  be  proved  that  L  -  ^  ^  ^,  or  ^  x  — ,  or  jB  x  — • 
^  F  F  F 

Since  F  contains  all  the  common  factors  of  A  and  B,  a  and  6  have  no 

common  factors  ;  consequently,  since  A  —  aF  and  B  =  bF, 

L  =  abF. 

Multiplying  by  F,  FL  =  abFF ; 

but  Ax  B  =  aFxbF=  abFF. 

Therefore,  Ax.  1,  FL  =  Ax  B, 

and  L  =  Aii^  or  ^  x  — ,  or  ^  x  — • 

F  F  F 

Proof  of  Principle  2. 

Let  L  be  the  lowest  common  multiple  of  A  and  B,  and  ilf  the  lowest  com- 
mon multiple  of  L  and  a  third  expression  C 

It  is  to  be  proved  that  M  is  the  lowest  common  nmltiple  of  A,  B,  and  C. 

Since  L  is  the  expression  of  lowest  degree  that  is  exactly  divisible  by  both 
A  and  B,  and  M  is  the  expression  of  lowest  degree  that  is  exactly  divisible 
by  both  L  and  C,  M  is  the  expression  of  lowest  degree  that  is  exactly  divis- 
ible by  ^,  B,  and  C 

Examples 

1.   Find  the  L. CM.  of  a?  +  6x?-\-llx  +  6  aii6.a^-4:x'-\-x-[-6. 

PROCESS 

Prin.  1,  L.  C. M.  =  (^  + 6.^  + 11. +  6)g-4a^  + .  + 6) 

^(x  +  l)(x'  +  5x  +  6)(x  +  l)(x' -  5x  +  6) 

x-\-l 
=  {x  +  l)(a^  +  5x  +  6){x'-5x  +  6) 

or  (x  +  l)(a:  4-  2)(x  +  S)(x-  2)(a;-  3) 

or  (a;  +  l)(a^-4)(a;='-9) 

Find  the  L.  C.  M.  of 

2.  4a3  +  7a2  +  10a-3  and  4a8  + Ga^  +  Ua +  3. 

3.  2a3-lla*  +  18a-14  and  2a'  + Sa"- 10a  +  14. 

4.  5a^-lla;*  +  3a;-l-12  and  5a;3-19ar'  +  27a;-12. 


132  LOWEST  COMMON  MULTIPLE 

5.  4a^- 14x2  + 22  a; -8  ^nd  2ci^ -3a^  -  x^ +  12x. 

6.  6a3  +  3a2-15a-75  and  2a»  + lla2  +  25a  +  25. 

7.  4a^-27a^-2a  +  15  and  2a* -9a^ -2Sa^ -15a. 

8.  3(f-llc'-32c-16  and  Sc^  -  19c2  +  8c  + 16. 

9.  4x*  -  7  a^  +  7  x"  ~llx  +  6  and  2a!* +  a^- a^- x- 6. 

10.  a;*-ar^  — 3a;  +  9  and  Sax* —  3  ax^ —  18  ax^ +  45 ax -27 a. 

11.  20a^  + 40^2^  25a; +  125  and  ea:^.^  7^^  ^  10x  + 25. 

12.  12m3-18m2  +  26m-10  and  15 Trv"  -  9 iri"  +  19 m -{- 10. 

13.  6  a^a;  —  5  a^x  —  18ax  —  Sx  and  6  a^b  —  13  a^b  — 6  ab  +  S  b. 

14.  4a;3  +  4a!2y-5a;2/='  +  252/3  and  4  a;3_  16^^.2^  ^25  a;/ -25 2^. 

15.  10a^  +  29a^-36a  +  9  and  Sa^ -\-34:a^ +  9a -9. 

16.  4  a;^  —  17  x^/  +  42/*  and  2x*  —  a^y  —  3  a^y^  —  5xy^  —  2y*. 

17.  5x*+8a^-27 x'+Ux-lO  and  3a;*+4a;3_i7a^_,_14a,_10. 

18.  2 a;*-9 ar' + 18 .^•2- 18 .T  + 9  and  3.T*-lla;3^17a^_12a;+6. 

19.  3a*+13a='-19a2+12a-4  and  4a*  +  22a3-2a2  +  2a  +  4. 

20.  63^  +  5x-6,  8a;2^i0a;-3,  10a;2_^9^_9 

21.  x*-2c(^  +  a^-l,  x*-x'-\-2x-l,  x*-3a^  +  l. 

22.  a;*-7a;2  +  9,  a;*  +  2  a;^  _^  ar^  -  9,  a;*- a;^  -  6a;- 9. 

23.  X*- 42^  +  40^-16,  a;*-12ar^+16,  a;* - 4 ar^  +  16 a;  - 16. 

24.  a;*-4ar''  +  4a;2_25^   a;*  -  4  a;^  _^  20  a;  -  25,  a;*  -  14  a;^  _,_  25. 

25.  4a;*  +  5a;2-a;-l,  6a;*  +  a;3_|.8ic2_l,  36^*-13a;2^1 

26.  10a;*+7a;3-33a;2^26a;-10  and  2a;*+7a.'3+5ar'-4a;-10. 

27.  16a;*  +  16ar'-48a;2-36a;  +  27     and     24  a;*  +  20  a;^  -  74 a;* 

—  45  a;  +  45. 

28.  10a;*  +  7a;«  +  2.T2-a;-2  and  6a^  +  5a^  +  4x +  1. 

29.  5a;*  +  3af*  +  r)ar'  +  a;  +  3  and  15 a;^  ^  ^4 ^.2  _j_  ^.  ^  j^2. 

30.  2a.'3-a;2_3a._,_2^  4a;3_^  6a^_  2a;_  4,  4a^-5x  +  2. 

31.  a;3-l,  2af +  2ar'-5a;  +  l,  a;3-3a;  +  2. 


FRACTIONS 


158.  A  fraction  is  expressed  by  two  numbers,  one  called  the 
numerator,  written  above  a  line,  and  the  other  the  denominator, 

written  below  the  line.     Thus,  -  is  a  fraction. 

6 

If  a  and  b  represent  positive  integers,  as  3  and  4,  the  fraction 
-  is  equal  to  - ;  that  is,  it  represents  3  of  the  4  equal  parts  of 

anything.     This  is  the  arithmetical  notion  of  a  fraction. 

But,  since  a  and  b  may  represent  any  numbers,  positive  or 

negative,  integral  or  fractional,  rational  or  irrational,  -  may  repre- 

4  ^ 

sent  an  expression  like  — .    Since  a  thing  cannot  be  divided  into 

of 
5§  equal  parts,  algebraic  fractions  are  not  accurately  described  by 

the  definition  commonly  given  in  arithmetic.  But,  since  an  ex- 
pression like  ^^',  regarded  as  20  fourths,  is  equivalent  to  5,  or 
20  -j-  4,  it  is  evident  that  the  numerator  of  a  fraction  may  be  re- 
garded as  a  dividend,  and  the  denominator  as  its  divisor ;  and  this 
interpretation  of  a  fraction  is  broad  enough  to  include  the  fraction 

Y  when  a  and  6  represent  any  numbers  whatever.     Hence, 

The  expression  of  an  unexecuted  division,  in  ivhich  the  dividend 

is  the  numerator  and  the  divisor  the  denominajtor,  is  an  Algebraic 

Fraction. 

The  fraction  -  is  read,  '  a  divided  by  6.' 
b 

159.  The  numerator  and  denominator  of  a  fraction  are  called 
its  Terms. 

160.  An  expression,   some  of   whose  terms  are  integral  and 

some  fractional,  is  called  a  Mixed  Number,  or  a  Mixed  Expression. 

a  —  b     r^  /j2  1 

c  — ,  —  —  2  -t-  — ,  and  a  —  b  -\ —  are  mixed  expressions. 

c        a^  x^  ab 

133 


134  FRACTIONS 


REDUCTION  OP  FRACTIONS 

161.  The  process  of  changing  the  form  of  an  expression  with- 
out changing  its  value  is  called  Reduction. 

162.  To  reduce  fractions  to  higher  or  lower  terms. 

163.  A  fraction  is  in  its  Lowest  Terms  when  its  terms  have  no 
common  divisor. 

1  X  3  a; 

164.  1.    How  many  eighths  are  there  in  -?   in  -?  in  — ? 

.     5a6o   •    4iCn   .    9ar>   •    8a;Vo 
in ?   m — ?   in—-?   in — ^? 

4  16  24  32 

2.  How  many  tenths  are  there  in  -?   in  —  ?   in  — ^? 

2  5  20 

3.  If  a  dividend  is  multiplied  by  any  number,  as  2,  and  the 
divisor  is  multiplied  by  the  same  number,  how  is  the  quotient 
affected  ? 

4.  If  a  dividend  is  divided  by  any  number,  as  2,  and  the  divisor 
is  divided  by  the  same  number,  how  is  the  quotient  affected  ? 

5.  Since  a  fraction  may  be  regarded  as  an  indicated  division, 
what  may  be  done  to  the  terms  of  a  fraction  without  changing 
the  value  of  the  fraction  ? 

165.  Principle.  —  Multiplying  or  dividing  both  terms  of  afrao' 
tion  by  the  same  number  does  not  change  the  value  of  the  fraction. 

The  proof  of  the  principle  is  as  follows : 

Let  a  and  b  be  any  two  numbers,  a  the  dividend,  b  the  divisor,  and  - 
the  quotient.     Also,  let  m  be  any  number. 

It  is  to  be  proved  that  «:=!??:«. 
0      mb 

Since  the  quotient  multiplied  by  the  divisor  equals  the  dividend, 

?  X  6  =  a.  (1) 

b 

Multiplying  (1)  by  m,  Ax.  4,  -  x  mb  =  ma.  (2) 

Dividing  (2)  by  mb,  Ax.  5,  q^m^,  ^8) 

0     mo 


FRACTIONS  135 

Hence,  the  terms  of  any  fraction,  as  -,  may  be  multiplied  by  any  num- 

6 

ber,  or  the  terms  of  any  fraction,  as  — ,  may  be  divided  by  any  number, 

mb 

without  changing  the  value  of  the  fraction. 


Examples 

1.  Reduce  — - —  to  a  fraction  whose  denominator  is  o?  —  bK 

a  +  b 

PROCESS 

(a2  _  62)  -4-  (a  +  &)  =  a  -  6 

a     _       a(a  —  b)       _a^  —  ab 
•*•    a  +  b  ~  {a  +  b){a-b)~  a?-b^ 

Explanation.  —  Since  the  required  denominator  is  {a  —  b)  times  the 
given  denominator,  both  terms  of  the  fraction  must  be  multiplied  by  (a  —  6) 
(Prin.). 

2.  Reduce  ^^^-^ — ^  to  its  lowest  terms. 

30  a^osz 

Explanation.  —  Since  a  fraction   is   in   its  lowest 
PROCESS  terms  when  its  terms  have  no  common  divisor,  the  given 

2  „  fraction  may  be  reduced  to  its  lowest  terms  by  remov- 

—    '  J  = --L       ing  in  succession  all  common  divisors  of  its  numerator 

oO  a  xz       10  az       ^t^^  denominator  (Prin.),  as,  3,  a,  a,  and  x  ;  or  by  divid- 
ing the  terms  by  their  highest  common  divisor,  3  d^x, 

3.  Change  —  to  a  fraction  whose  denominator  is  4  61 

u  0 

5  a 

4.  Change  —  to  a  fraction  whose  denominator  is  42. 

5.  Change to  a  fraction  whose  denominator  is  656. 

3a 

6.  Change to  a  fraction  whose  denominator  is  84  xy. 

14  a; 

7.  Change to  a  fraction  whose  denominator  is  20  m'. 

62/ 


136  FRACTIONS 

„  q 

8.  Change  to  a  fraction  whose  denominator  is  {x  —  1)'. 

ar  —  1 

9.  Change  ''^~ —  to  a  fraction  whose  denominator  is  (2  x-\-by. 

2x-\-5 

10.  Change  — - —  to  a  fraction  whose  numerator  is  3  a  +  a}. 

o  —  a 

11.  Reduce  ^  ~     to  a  fraction  whose  denominator  is  a?  —  b\ 

a  +  b 

12.  Eeduce  — — ^  to  a  fraction  whose  numerator  is  ic^  —  y\ 

2x  +  y 

13.  Reduce to  a  fraction  whose  denominator  is  a  —  6. 

b  —  a 

14.  Reduce  -^^^ —  to  a  fraction  whose  denominator  is  4  —  oj^. 

x-2 


Reduce  the  following  to  their  lowest  terms : 

15.   ^^.  22.    -^^^y"^'- 

a^xy  —  100  x*y^ 

am^n*  x'^y* 

o?b^3(?  _.     xT-^a? 

Wxy"^  »"• 

16  mhiy?z^  x"+^y 

40  am^ya^  xy"*'^^ 

jg     210&c^d  26     ^"""^^ 

750  a&^c'  ■      ax 


42  ab^cd*  3  a^^ft 

121  a^b^c'  '     2ar]f 


FRACTIONS  137 


„„     ax"-»+i                                _     a" -11a +  24. 
29.    —  45. — — • 

ftaj"-"  a^  —  a—6 


n(n-2)ab  '   a^  +  2x'~35x 

31           a'-6'  4-     7a;-2g^-3 

a2  +  2a6  +  &2'  '   2ar'  +  7a;-4' 

g'^  -  2  a6  +  ft''  a(a  +  2  6)* 

a2_52       *  •    6(a2_4j2^2 

8  a' +  27  ar*'  '   a*  -  2  g'^ft^  +  ^^4* 

,^     3a*  +  3g6  ^-     a^-2x*  +  a^ 

o4.     — •  50. • 

g*  +  ab^  x^  —  or 

3  x'y  —  6xy  -     af-{-5ot^  —  6x 

o^y  -8xy'  '         2x^-2 

oo. — — -•  5<o. 


2g3&-2&*  '    x*-10ar'  +  9 

4  g^  -  ab^  ar'  -  21  a;  +  20 

8a*  +  g6«'  *    a^*-26a^  +  25' 

3g     2a^y^-8y*  ^^     a^  +  3a^  +  3a;  +  l 

'    4:0!^y  —  32y*  '     a^  +  x^  —  4a;  —  4 

a*bc  -  b'c  a?  -  3  a?b  +  3  g6^  -  W 

'    '6a%-\-3V'  '            3a'b-3ab^ 

10  7ia;  +  10j«v  _„      3  g- 4-4  ga;  —  4a;^ 

40.     '-^'  5d.     • 

25  iix"  -  25  ny^  9  g"  -  12  ga;  +  4  x^ 

2  ga;  —  gy  —  4  6a;  4-  2  6y 
4  ga;  —  2  gy  —  2  6a;  +  6y 

^„       g"+^-gY  >o          9.r3-13g2a;-4g« 

42. — •  58. 

a»+3  +  a"+'/  3  6a;  +  3  a;y  —  4  g6  —  4  gy 

a;*y  +  x^y^  +  y°  gg     m  —  m^  —  n-\-  mn 

a;6  _  y6  ^  _  ^^  ^  ^2  _  j^ 

. .     x*y  —  a;^y^  +  y^  gm  —  g/i  —  m  +  ?^ 

a;®  4-  J/*  am  —  an  +  m  —  n 


™n+2         ^n 

41.    ±_ ±..  57. 

/j»w4-3      ^_  /j»H 


138  FRACTIONS 

61.   Keduce  -— — — ^   ^  '     -—  to  its  lowest  terms. 

Solution 

The  process  of  finding  the  H.  C.  D.  of  the  terms  of  the  fraction  can  be 
shortened  in  some  instances  by  finding  the  sum  or  the  difference  of  the 
terms,  since  the  result  will  either  be  the  H.C.D.  or  some  multiple  of  it,  §  148. 

3  a:3  -  16  x2  +  25  a:  -  12 
3  a:3  -  8  a;2  -  7  a;  +  12  * 

Subtracting  the  numerator  from  the  denominator, 

8  a;2  -  32  X  +  24 

=  8(a;2-4a;  +  3). 

By  trial,  a;2  -  4  x  +  3  is  found  to  be  the  H.  C.  D. 

Dividing  the  terms  of  the  fraction  by  x"^  —'ix-\-  Z,   the  fraction  in  its 

lowest  terms  is -• 

3x  +  4 

Reduce  the  following  to  their  lowest  terms : 
a,-3  +  5a^-9a;-45 


62. 


63. 


64. 


65. 


66. 


67. 


68. 


69. 


0^  +  3x2-25  a;-75 

a^  +  2a^-23a;-60 
:i?-llx'-10x  +  200* 

4,a?j^qx'  +  l{)x-^ 

x'  +  ^x'  +  ^x+Q 
a,-3  +  3a^  +  4a;  +  2' 

5af-17x2  +  16a;-4* 

5x^-  14a^  +  22a;  +  5 
5a^-18ar'  +  34.'B-15* 

a^  +  6  3i?y  —■  2  xy^  —  5  y* 

g"  +  6^  4-  2  c^  +  2  a6  +  3  ac  +  3  6c 
a'  +  b'  +  c'  +  2ab  +  2ac-\-2bc' 


70. 


71. 


72. 


73. 


FRACTIONS  139 

^2  +  62  -I-  c2  -f  2  a6  -  2  ac  -  2  6c 


a2  +  62. 

-  c^  +  2  a6 

a2  -1-  62  +  c^  - 

■  2  a6  —  2  ac  +  2  6c 

a2  +  62  +  5  c^  - 

-  2  a6  -  6  ac  +  6  6c 

4a2  +  9  6^  +  16  c2  +  12a6  +  16ac 

+  24  6c 

4a2-962  +  16c2  +  16ac 


ab  (x^  —  y^  +  xy{d^  —  6^ 

166.  Signs  in  fractions. 

167.  The  sign  written  before  the  dividing  line  of  a  fraction  is 
called  the  Sign  of  the  Fraction. 

It  belongs  to  the  fraction  as  a  whole,  and  not  to  either  the 
numerator  or  the  denominator. 

In  — —  the  sign  of  the  fraction  is  — ,  while  the  signs  of  x  and  3  z  are  + . 

168.  An  expression  like  ^^  indicates  a  process  in  division,  in 

—  6 
which  the  quotient  is  to  be  found  by  dividing  a  by  6  and  prefixing 
the  sign  according  to  the  law  of  signs  in  division ;  that  is, 

—  a,a  -\-a      ,  a 

—  6         6  +6  6 

—  a_     a  +  a_     a 
+  6~      6'                         -b~     b 

By  comparing  the  above  fractions  and  their  values  the  following 
principles  may  be  deduced : 

169.  Principles.  —  1.  TJie  signs  of  both  the  numerator  and  the 
denominator  of  a  fraction  may  be  changed  without  changing  the  sign 
of  the  fraction. 

2.  Tlie  sign  of  either  the  numerator  or  the  denominator  of  a  frac- 
tion mmj  be  changed,  provided  the  sign  of  the  fraction  is  changed. 

When  either  term  of  a  fraction  is  a  polynomial,  its  sign  is  changed  by 
changing  the  sign  of  each  of  its  terms.  Thus,  the  sign  of  a  —  6  is  changed 
by  writing  it  —  a  +  6,  or  6  —  a. 


140  FRACTIONS 

Examples 
Reduce  to  fractions  having  positive  numbers  in  both  terms : 
—3         „     —a—x         _  —a—h         „         —2—n 

o. -— •  /. 


-4  2x  c-\-d  2  +  n 

2.    -^.        4.       -4^.         6. ^2_.        ^     -4(a  +  &). 

-5  -b  —  y  —d  —  y  5{—x  —  y) 

170.  Since,  from  the  laws  of  signs,  changing  the  signs  of  an 
even  number  of  factors  does  not  change  the  sign  of  the  product,  it 
follows  that : 

Principle  1.  —  The  signs  of  an  even  number  of  factors  of  the 
numerator  or  of  the  denominator  may  be  cJianged  without  changing 
the  sign  of  the  fraction. 

Since,  from  the  laws  of  signs,  changing  the  signs  of  an  odd 
number  of  factors  changes  the  sign  of  the  product,  it  follows  from 
Prin.  2,  §  169,  that : 

Pbinciple  2.  —  The  signs  of  an  odd  number  of  factors  of  the 
numerator  or  of  the  denominator  may  be  changed,  provided  tht>  sign 
of  the  fraction  is  changed. 


Examples 

1. 

Show  that 

-b          b 
b—a     a—h 

2. 

Show  that 

—  a                a 

6  —  a  +  c     a  —  h—c 

3. 

Show  that 

2                     2 

a{b  —  a)         a(a  —  b) 

4. 

Show  that 

1                               1 

{a-b){c-h)         (a-b)(b-c) 

5. 

Show  that 

m  —  n                   m  —  n 

(a  —  c)(b  —  a)      (c  —  a)(a  —  b) 

6. 

Show  that 

1                                           1 

(6-a)(c-6)(a-c)         (a -&)(&- c)(c- 

-a) 

7. 

Show  that 

n—m                                m—n 

. 

(y  -x){z-  y)  {x  -z)      {x-  y)  (y  -z){z-  x) 


FRACTIONS  141 

171.   To  reduce  a  fraction  to  an  integral  or  a  mixed  expression. 

1.  How  many  units  are  there  in  ^-^-^   in  ^\^?   in  ^  ? 

2.  How  many  units  are  there  in ^ c    m r 

Examples 
1.   Reduce  ^^'^     to  a  mixed  number. 

X 

PROCESS  Explanation.  —  Since,  §  158,  a  fraction  may  be  re- 

,  ,      garded  as  an  expression  of  unexecuted  division,  by  per- 

"*      =z  a  -\ —     forming  the  division  indicated  the  fraction  is  changed 

^  ^     into  the  form  of  a  mixed  number. 

"When  the  degree  of  the  numerator  is  lower  than  the  degree  of  the  denomi- 
nator, the  fraction  cannot  be  reduced  to  an  integral  or  mixed  expression. 

Reduce  the  following  to  integral  or  mixed  expressions : 

2. 


3. 
4. 

5. 
6. 

7. 


a?- 

3a^4-a-l 

a' 

4a^ 

-Sx^  +  2x- 

1 

2x 

ah  - 

-bc-cd  +  cP 

b 

aV 

—  ax^  —  x  —  1 

ax 

x"- 

x-15 

X 

-4 

c^- 

2xy-f 

10. 


11. 


1'> 

a;3  4.5ar!  +  3a;-6 

a;  +  2 

13. 

a'  +  9  a^  +  24  a  +  22 

a  +  3 

14. 

ar5_6a:2  +  14a;-9 

x-1 

15. 

cB*-3a^  +  5a;-l 

x-Z 

16. 

a*  -f  3  a^b""  +  &* 

a2  +  &' 

17. 

4  a;2  +  22  a;  +  21 

2a;  +  4 

18. 

a^-3a^  +  4a;-3 

a;  —  4 

19. 

a^  +  2  a&  +  6^  +  c^ 

a  +  6  +  c 

f>(\ 

a3-6a26  +  l2a&2-10  6' 

a^  —  6  xy  +  4  y^  ^ 
2xy 

a2_2a-26 
a  +  5 

a«  +  2  a&  +  h\ 

a-\-h  a  —  26 

x^  +  f  21.    a;^  +  4a:^y  +  6a^/  +  4a;y^^ 

x'-xy  +  f  '  x-\-y 


142  FRACTIONS 

172.   To  reduce  an  integral  or  a  mixed  expression  to  a  fraction. 

1 .  How  many  fifths  are  there  in  6  ?  in  10  ?  in  a  ?  in  3  6  ? 

2.  How  many  fifths  are  there  in  64  ?  in  a  +  — -  ?   in  2  a  +  -  ? 

5  5 

Examples 

1 .   Reduce  a  +  -  to  a  fractional  form, 
c 

PROCESS  Explanation.  —  Since  1  =  c  -h  c,  a  =  ac-^c. 

Since,  §  158, 

_ac  I,  5 

<*  —  "T  -  means  6-^c,  a  +  -  =  ac-^c  +  6-HC. 

c  c  c 

b  _ac     b  _ac-\-b    §  104,  3,  =  (ac  +  b)^c. 

Rule,  —  Multiply  the  integral  part  by  the  denominator  of  the 
fraction;  to  this  product  add  the  numerator  when  the  sign  of  the 
fraction  is  plus,  sid)tract  it  when  the  sign  of  the  fraction  is  minus, 
and  write  the  result  over  tJie  denominator. 

If  the  sigji  of  the  fraction  is  -,  the  signs  of  all  the  terms  in  the  niunera- 
tor  must  be  changed  when  it  is  subtracted. 

Reduce  the  following  to  fractional  forms : 

a^  —  db 


2. 

-1- 

3. 

._|. 

4. 

5c  +  — 

c 

5. 

36-^ 

+  26^ 
b 

g 

4a; -1-''^ 

—  X 

7. 

40:+     ^ 
,  ,    ,  2  6c  -  1 

8.    a  — 


9.   a  — 


10.    fe- 


ll,   a 


b 

a  —  b—c 
2 

b  —  a  —  c 
2 

c  —  a  —  b 


12.  ^L±A±i_c. 

13.  6  +  ^'+^^ 

ab  —  6* 


a  —  X 

a^  +  4: 

X-4: 

A.h2         _ 

W 

FRACTIONS  143 

14.  a  +  x 17.    ^•\-xy  +  if-\ — 2l — 

"     ~  x-y 

15.  x  +  5--^^^r^-  18.   3a-6&-^^^'-^^. 

a+2h 

16.  a^-ah  +  b^ " —  19.    x-a?-^ ~. 

a+b  1+x       ■ 

173.  To  reduce  dissimilar  fractions  to  similar  fractions. 

1.  Into  what  fractions  having  the  same  denominator  may  |, 
|,  and  f  be  changed  ? 

12  4 

2.  Express  -— ,  — -,  and  -—-  by  fractions  whose  common  de- 

sx  5x  15a; 

nominator  is  the  lowest  common  multiple  of  the  given  denomi- 
nators. 

174.  Fractions  that  have  the  same  denominator  are  called 
Similar  Fractions. 

175.  Fractions   that  have  different   denominators  are  called 
Dissimilar  Fractions. 

176.  Principle,  —  The  lowest  common  denominator  oj  two  or 
more  fractions  is  the  lowest  common  multiple  of  their  denominators. 

The  abbreviation  L.C.D.  is  used  instead  of  Lowest  Common  Denominator. 

Examples 

1.   Eeduce  - —  and  - — -  to  similar  fractions  having  their 
3bc  ooJ) 

lowest  common  denominator. 

PROCESS  Explanation.  —  Since  the  L.  C.  D.  of  the 

f>  n    2        given  fractions  is  the  lowest  common  multi- 

—7-  =  —-- ~  =  pie  of  their  denominators  (Prin.),  the  lowest 

60c      3bc  X  ^a      b  abc      common  multiple  of  their  denominators  must 

c     __     c  X  c     _     (^  be  found.     This  is  6  abc. 

6ab  ~  6ab  X  c  ~  6  abc  '^^  reduce  the  fractions  to  equivalent  frac- 

tions having  the  common  denominator  6  abc, 
§  165,  the  terms  of  each  fraction  must  be  multiplied  by  the  quotient  of  6  abc 
divided  by  the  denominator  of  the  fraction. 


144  FRACTIONS 

Rule.  —  Find  the  lowest  common  multiple  of  the  denominators 
of  the  fractions  for  the  lowest  common  denominator. 

Divide  this  denominator  by  the  denominator  of  the  first  fraction, 
and  multiply  the  terms  of  the  fraction  by  the  quotient. 

Proceed  in  a  similar  manner  with  the  other  fractions. 

All  mixed  expressions  should  first  be  reduced  to  the  fractional  form,  and 
all  fractions  to  their  lowest  terms. 

Reduce  to  similar  fractions  having  their  L.  C.  D. : 

2.  ^  and  ^. 

2  6 

„    ab       1  a? 

3.  —  and  — 

4  y 

4.  «  and  ^. 

3  6y 

2a        ■,  3x 

5  6  4  a  a^y*'    a^y^'    a^ 

6.  ^  and  ^-  12.    ^±i^,    ^  ^ "  y,    '^.^. 
c'd           cd^           .  2  4  6 

7.  -1-  and  ^.  13.    ''-^=^,    ^^^,    a-^ -^ 
2xy           4ay  x  y  3x 

14. 


15. 


8. 

m  —  n 
a 

2        *     . 
m-\-  n 

9. 

4  6c 

3ac 

7a6 

3aV 

4  6V 

6ac2 

0 

3a6 

8aV 

la" 
^l)'c 

5 

a36c2 

.1. 

3 

-6 

3 

16. 


17. 


18. 


19. 


13.    - 

4  ay 

X 

-5                        ■) 

y 

01?                X                X 

or^  —  l'    a;  +  l'    x  —  \ 

a^            a           2  a 

a* -16'    a2  +  4'    a=^-4 

4  a         3  6           1  "^^ 

a  -  6'    6  +  a'    a^  -  ? 

a              X           —ax 

1  —  ax'    1  +  ax    ax  + 1 

1                      1 

^ 

1 

x'  +  lx-^lQ'    x'  +  x-2' 

h4a; 

a+5                a-2 

a  -,  X 

a2_4a  +  3     a'^-Sa+lS     a^-ea+S 


FRACTIONS  145 


ADDITION  AND  SUBTRACTION  OP  FRACTIONS 

T.-  J  XI-        1        „  4     3     7     2     a     &     1     2 

177.  1.    Find  the  value  of  -  +  -; :    -+    :    -  +  — 

X     X     a     ay     y     x     y 

2.  What  kind  of  fractions  can  be  added  or  subtracted  without 
changing  their  form  ? 

3.  What  must  be  done  to  dissimilar  fractions  before  they  can 
be  added  or  subtracted?  How  are  dissimilar  fractions  made 
similar  ? 

178.  Principle.  —  Only  similar  fractions  may  he  united  by 
addition  or  subtraction  into  one  term. 

Examples 

1.   Find  the  algebraic  sum  of  -  + 

b     b     b 

I 

PROCESS 

J.  ^A^   o  a,cda  +  c  —  d 

§104,3,  -  +  ___  =  ___ 


2.   What  is  the  sum  of  —,    — ,   and    ^? 
4  '    10  12 


PROCESS 

Zx  ,  7x  ,  5v 

— - 

4       10      12 

^\.,x     42a;     25y 

^        60        60 

^87a;  +  25y 

-  60  \ 

Explanation, — Since  the  fractions  are  dissimilar,  they  must  be  made 
"'IT  befotP-they  can  be  united  into  one  term  (Prin.). 

•^^mmon  denominator  is  60.    3x  ^  45*     7*  ^  42^     5j/ ^  25]/ 
,   ^  4        60       10       60       12       60 

Therefore;  the  sum  is  ^^J  +  Ji^  +  gil  =  ^5x  +  42  x  +  25y  ^87  x  +  25y. 
60        60        60  60  60 

ALG.  10 


146  FRACTIONS 

3.    Imd  the  algebraic  sum  of 

^  8  7  4 

PROCESS 

bx-1     3a;-2     a;-5^35a;-7      (24a;-16)      14a;-70 
8  7  4  56  56  56 

^  35  a;  -  7  -  (24  a;  -  16)  +  14  a;  -  70 

56 
^35a;_7_24a;4-16  +  14a;-70 

56 
^25a;-61 

m 

Suggestion.  —  When  a  fraction  is  preceded  by  the  sign  — ,  it  is  expedient 
for  the  beginner  to  inclose  the  numerator  in  parenthesis,  if  it  is  a  polynomial, 
as  is  shown  above. 

Rule.  —  Reduce  the  fractions  to  similar  fractions  having  their 
lowest  common  denominator. 

Change  the  signs  of  all  the  terms  of  the  numerators  of  fractions 
preceded  by  the  sign  — ,  then  find  the  algebraic  sum  of  the  numera- 
tors, and  write  it  over  the  common  denominator. 

1.  Reduce  the  resulting  fraction  to  its  lowest  terms,  if  necessary. 

2.  The  integral  and  fractional  parts  of  mixed  expressions  may  be  united 
separately. 

3.  An  integer  may  be  expressed  as  a  fraction  whose  denominator  is  1. 

Find  the  algebraic  sum  of 
.     a—b,b—c 

4. 1 • 

ab  be 

_     a  +  b  ,  a  —  b 

o. 1 • 

a  —  &     a  +  b 

6.  -J^  +  a. 
a  —  x 

7.  a  +  6  + 


a  —  b 


9. 

b-c 

a  —  c 

be 

ac 

10. 

a-\-b 

a  —  b 

a  —  b 

a-hb 

11. 

5a' +  b 
a'-b' 

2 

--2. 

12. 

x  +  y- 

o^-hy'. 

x-y 

19 

X 

x-2 

8         a;  +  l  1     , 

*   x'  +  x  +  l     x-1  '   x-2     x  +  2 


FRACTIONS  147 

Simplify : 

2a;  +  la;-2     a;  —  3     5  —  a; 
3  4  6  2* 

x-2     x-4.     2-Zx     2x  +  l 
6  9  4  12     '        . 

x  —  1      X  —  2      4a;  —  3      1  —  X 
"    "3  18  2f~         6 

,„     2-6a;  ,  4a;-l      5x-3     1-x 
5  2  6  3 

a;  +  3      a;  — 2      a;  — 4      a;  +  3 
4  5  10  6    * 

a;  — 4     x  —  6  ,  ^      x-{-S 

„^     2-3a;     3-2a;  ,  l-4a; 

20.  — +  x 

3  4  o 

^,     l-2a,2a-l      2a-a2  +  l 

21. r 


22. 


5  4  8 

S  +  x-a^     1-x  +  x^      l-2a;-2a;' 
4  6  3  ' 


23.    §^  +  ^-3+i.  30.       ^-&    +^+^'         ^ 


5       2  6  2(a+6)      a^-fi^     a-b 

„.     a  —  ba  +  b,6ab              _,      a-f-1          6       ,10 
24.   — — 7+-^^ — n-  31.    — — — P  + 


a  +  6      a  —  6     a^  —  b^  o.^  —  9     a  +  5      a  +  3 

25.  a±x^a-x_^_2a^^  32_    2  a  -  3  6  -  i^^^±^'. 
a  —  a;a4-a;a^  —  a;^  2a  +  3b 

,  i    ,  ^  —  ^  ooQ         o        8a^  — 4a; 

26,  a;4-lH -•  33.    3a  — 2x  —  — -— • 

a;  —  1  6a-\-  2x 

^^     ^_ax-bx  +  ab  34_    m-^'  +  ^^'  +  n. 

a;^  m  —  w 

28           g  +  l        ■        Ct-l  35      _J: 1___l1. 

•    a^  +  a  +  l      a'-a  +  l  '    2(a;-l)      2(a;  +  l)^ar' 

29.    3a;+  — -f2a;  +  AY  36.    ^4-l4--l^-2. 

ax     V          <*^/  ^            1  +  a-' 


148  FRACTIONS 

07        O'         ft  — 2  ,       3 


a  —  2      a  +  2     4  —  a- 
SuGGESTiON.— By  Prill.  1,  §109,       ^  -3 


4  -  a2     a-'  -  4 


38.   5L±1  +  ^^JZ1+    2a'' 


a  —  1     a  +  1      1—  a' 
89.  «i±2+     2  3 


£c2-4a;-2     2-a; 

^^    x(a  +  x)     3ax-af  ^  ^^ 
a  —  X  X  —  a 

41     cf  +  &     g^  +  &".&  —  g 
g  —  6      b^  —  a^     a  +  b 

42.    ^L^  +  ^^  +  _i 

a;  (a;  —  g)      g  (g  —  x)      x  — 

43.    1-1^+.  1 


g3  +  8      8-g3     4-g-^ 
a;  —  1        ,       a^  3 

44.     ^; — r  + 


x^  —  5x  -\-6      X  —  2     x  +  1 
_     2  ,        3  2a;-3   ,        1 


X     l-2a;      43^*-!      l  +  2a; 
..a^  +  g^,        a;  +  g  2 

4o.     — --J-- 


•3?  —  c^     a;^  +  ga;4-g^     x- 
.„  3  m  771  +  2  a; 


48. 


(m  —  2  x)^      (m  +  .t)  (m  —  2  a;)      m-\-x 

3 2 

if  —  m?/  —  12  tii^     y^  —  5 m?/  -\-^'ni^ 


._  g6  g?)  ^ 

49. 1. 

g2  -  g6  +  62     ^^  +  g6  +  6^ 

50.      „       1         „-    .       1         „+  1 


51. 


52. 


x'2-3a;  +  2     a^  +  2a;-3     a;2  +  a;-6 
2  11 


ar'  +  SaJ  +  e      a;' +  6  a; +  8      a;2  +  7a;4-12 
5(a;-3)  2(a;  +  2)  x-\ 


a^-x-2      x-2  +  4a;  +  3     6-a;-a^ 


FRACTIONS  149 


_„     a  —  h  —  c     a  —  h  +  c.  4rtc 

5o.      — ; ; h 


a  +  h  -{-  c     a  +  6  —  c      (a-\-hy  —  c^ 
54.    Simplify     „        — —z:—2  +  —» — 7. — —7' 


+  1  a'  +  2a  + 

Solution 


ff -^  -  2  a  +  1  a^  +  2  a  +  1      \        a^  -  2  a  +  1  /  \        a^  +  2  a  +  1  / 


4a  4a  16a^ 


a2-2a+l     a2+2a+l     (a-l)2(a+l)2 

Suggestion.  — Frequently,  by  reducing  one  or  more  of  the  given  fractions 
to  mixed  numbers,  the  integers  cancel  each  other  and  the  numerators  are 
thus  simplified. 

Simplify : 

55. 

56. 

57. 

58. 

59. 

60. 

61. 

a  —  b     a  +  b     a^  +  b^     a* -\- b* 

Solution 
Combining  first  two  fractions,  -^ ^  =    ^  "^  .      (1) 

Combining  (1)  with  the  third  fraction,     -^^ ^^^  =   ^"^^  •      (2) 

Combining  (2)  with  the  fourth  fraction,  -^^ i^  =A^^.     (3) 

Hence,  «  «  ^  «*         4  a68   _   8a67 


a2+62 

1+    ^<'b  . 

a^  +  3ab  +  2b^ 

a"-  13  b^ 

a'-^Sab-Ab' 

a'-16lr' 

x^-\-x  +  l       x" 

—  x-{-l 

af-x  +  1     a^- 

-2a;+l 

a^  +  a^  +  x  +  1 

1         ^ 

a^  —  a^  +  x  —  l 

^     x-1 

X  +  1      x—1 
X  —  1      x  +  1 

x+2     x-2 

x-2     x+2 

x-{-3     x~3 
x—3      x+3 

X-\-4:          X  —  4: 
X  —  4:          X+  4: 

a            a 

2ab         4.ab^ 

a-b     a  +  b     a^  +  62     a*  +  b*     a^  -  6" 


150  FRACTIONS 

-„     a  +  h     a  —  b       4  a6     ,     Sab^ 
D<6. ;;  ~r 


a-b     a  +  b     a'  +  b^     a*+b* 


a-b     a  +  b     a^ -{- b^     a* -\- b* 

_ .         1       ,  _1 x__ __ 7?  +  ^X 

'   x  —  1      x  +  l      x^  +  1       0^"  +  !' 

a-\-x     g^  +  a^  _a  —  a3_a^  —  a^_4  a^x  -}-  4  aa? 
a  —  a;     a^  —  x^     a  +  x     a^  +  x^  a*  —  x* 

66.    Simplify -^ +  . 


(6  —  c){c—  a)      {a  —  c)  (a  —  6)      (b  —  a)(c  —  b) 


Solution 
a  b 


+  ■ 


(6  — c)(c  — a)      (a  —  c)(ffl  —  6)      (6  — a)(c  — 6) 
§  170,  = ^^ + ^^ +  *" 


(6  —  c)(c  —  a)      (c  —  a)(a  —  b)      (a  —  6)  (6  — c) 

_a(a—  b)+  b(b  —  c)4-  c(c  —  a)_ 
(6  —  c)(c  —  a)  {a  —  b) 


Simplify : 


67.    = \ 1 . 

(&  —  c)(a  —  c)      (c  —  a)  (a  —  b)      (b  —  a)  (6  —  c) 

68.  «  +  l  I  b  +  1  c  +  1 


69. 


70. 


(a  —  6)  (a  —  c)      (6  —  c)  (6  —  a)      (a  —  c)  (6  —  c) 

c^a&  &^ca  a^bc 

(c  —  a)  (6  —  c)      (6  —  a)  (6  —  c)      (a  —  b)(a  —  c) 

b  —  c c  —  a a  +  b 

(b  —  a)  (a  —  c)      {b  —  c){a  —  b)      (a  —  c)  (b  —  c) 


c  +  a                    b  +  c  ,  a  +  b 

71-    -. rrr. r  — 7 tt; r  + 


72. 


(a  —  b)(b  —  c)      (c  —  a)(b  —  a)      (c  —  b)(a  —  c) 

c  +  a c  —  b a  —  b 

{a  +  b){b  —  c)      (c  —  a)  (a  +  b)      (b  —  c)  (a  —  c) 


_  c  +  a  —  b       .       b  +  c—a  a  +  b  +  c 


(a  —  6)  (6  —  c)      (c  —  a)  (a  —  6)     (6  —  c)  (a  —  c) 


FRA  CTIONS  151 

MULTIPLICATION  OP  FRACTIONS 

179.  1.    How  much  is  5  times  f?   2  x  ^?   6x^?   3  x  — ? 

4x^?  3x^?   ex??    ax^? 
5  4  4  d 

2.  Express  5  x  ^^  in  its  lowest  terms ;  3  x  | ;  4  x  ^r^;  2  x  -— ; 

Q  K  7  12  6 

Uy—-   7  X  — •    10  X  — •   8  X 
^"^22'    ^14'   ^"^20'      "^16 

3.  In  what  two  ways,  then,  may  a  fraction  be  multiplied  by  an 
integer  ? 

4.  How  much  is  ^  of  \%  or  ^  ^  5  ?  i  of  ^,  or  —  --  4  ? 

5.  How  much  is  ^  of  |,  or  |  -=-  5  ?  ^^  of  -— ,  or =-  4  ? 

6.  In  what  two  ways,  then,  may  a  fraction  be  divided  by  an 
integer  ? 

180.  Principles.  —  1.  Multiplying  the  numerator  or  dividing 
the  denominator  of  a  fraction  by  any  number  multiplies  the  fraction 
by  that  number. 

2.  Dividing  the  numerator  or  multiplying  the  denominator  of  a 
fraction  by  any  number  divides  the  fraction  by  that  number. 


Examples 


1.    Multiply  I  by  5. 
0        d 


PROCESS  Explanation.  —  To  multiply  -  by  -  is  to  find  c  times 

b        d 
a     c__ac 

b      d      bd         *^®     P*'^  o*  -•    -  P*"^  of  -  =  —  (Prin.  2),  and  c  times 
d  b     d  b     bd 

-^  =  ^  (Prin.  1).     Therefore,  7  x  ^  =  ^• 
bd     bd^  ^  b     d     bd 

OB 

To  find  the  product  of  ^  x  |- 
0       u 

Let  »-ix|-  (1) 

0     a 


152  FRACTIONS 

Multiplying  each  member  of  the  equation  by  6  x  d,  Ax.  4, 

xxb  X d  =  -x-xb X d 
b     d 

§  104,  1,  =^xbx-xd  =  ac. 

b  d 

xbd  =  ac. 

Dividing  by  bd,  Ax.  5,  « =  —  •  (2) 

bd 

.-.  Eq.  (1)  and  (2),  Ax.  1,  ^x|  =  f§. 

b     d     bd 

Similarly,  «x:^x?  =  «^; 

b     d     f     bdf 

and  so  on  for  any  number  of  fractions. 

Rule.  —  Multiply  the  numerators ,  together  for  the  numerator  of 
the  product,  and  the  denominators  for  the  denominator  of  the 
product. 

1.  Cancel  equal  factors  from  the  numerator  and  denominator. 

2.  Reduce  integral  and  mixed  expressions  to  the  fractional  form  before 
multiplying. 

2.    Simplify  ^^±^  X  ^^^  X  —^ 

^    ^       x  +  3  x-\-5      'Jaf-lOx 

Solution  3x^  +  16x  ^  gj-g  ^  1 


X4-3  x  +  5      2x2_i0x 

^3x(x  +  5)  ^^  (x  +  S)(x-3)  ^^ 


x  +  S  x  +  6  2x(x-5) 

Canceling,        ^  3  (x  -  3) 

^  2(x  -  5) 

Multiply : 

3.    ^   bv  -^^  6     i^  bv    -^^^ 

4:xy     -^    3a^  Sxy     ^        IGm^ 


4     ^  bv  ^-^-  7     ^^  bv   -1^ 

'    2  ac     -^   10  ?/2  '    12  by     ^  x'  ' 

_     4a6    ,      3  6c  ,  _     a'"6"  ,         6.t« 


lOc^     -^     a»  4  a;     "^    a'"-^^*" 


FRACTIONS  153 

9.    <^'by^'.  n.    ^^  by  25-10 »:. 


b"'+^     'or  20 -8  a; 


x-y 


-_         a      ,  h  ,„1  —  or<;  +  6a^,2 


a-\-h         a  —  h  2—^x-\-x^         1—x 

Simplify  the  following : 

13.    (a-6)%.       b       ^Aa-hbY 
a  +  b       o?  —  ah       o?  —  b'^ 

14     ^*  ~  '^^  X  ^  +  ^  X  g^  —  aa;  +  ar^ 
a''  +  a:^     a^  —  o?         (a  +  a;)^ 

,_4a  —  &  2a  4a:^  —  ?/2 

15.    7i X— -X —^i- 

2x  +  y     Aa^  —  ab  4 

j6.  £±4  X  y  -  ^i  X  — ^ 


a; -3      2^2-8      j9x  +  3i> 
17.    i^'-g"  X  ^~^    X  -^— . 

18     a^  +  8      a'^  +  2a  +  4 
•    a3_8     a2-2a  +  4' 

jg     g*  +  g V  +  a;^  ^ 


20. 


g*  —  ga;*  g^  —  ga;  +  a^ 

g*  +  4      ^   g^  +  g  +  l 


g*  +  g^  +  1      g2  +  2  g  +  2 


21     g^  +  gx  +  e^^a^'  +  Ta^  +  lO 
ar'  +  Gaj  +  S      a^4-7a;  +  12 

g^  +  3  a;  -  10      g^  -  10  a;  +  21 

a^-4a;-21       ar'  +  Taj  +  lo' 

g^4-g^+gc4-?>c      a^—ax-{-ay—xy     x^—x{y—d)—ay 
ax~ay  —  x^-\-xy      a^-\-ac+ax+cx     a^—a(j/—b)—by 

a^-5x^  +  8x-4:       a;«  -  10  ar^  +  33  a;  -  36 
•   a:3_8ar'  +  19x-12       a^-Gx'  +  llx-G  ' 

a^-3ar^-23a^  +  75a;-50        a^- lOar' +  29a;- 20 
'    aj*-oa;«-21a;2  +  125a;-100      a;^  _  ^2  a;^  +  45  a;  -  50' 


164  FRACTIONS 


DIVISION  OF  FRACTIONS 

181.  1.  How  many  times  is  \  contained  in  1  ?    ^  in  1  ?    ^  in  1  ? 

2.  If  the  numerator  of  a  fraction  is  1,  how  does  the  number  of 
times  the  fraction  is  contained  in  1  compare  with  the  denominator  ? 

3.  How  many  times  is  -  contained  in  1  ?     in  1  ? 

d  x  +  y 

o  _ 

4.  How  many  times  is  ^  contained  inl?     -inl?     -inl? 

d  d 

182.  The  reciprocal  of  a  fraction  is  the  fraction  inverted. 

For,  since  -  multiplied  by  -  produces  1,  §  102,  if  1  is  the  dividend  and 
a  b 

-  the  divisor,  -  is  the  quotient. 
h  a 


Examples 


1.   Divide  -  by  — 
b     ^  d 


Explanation. — The  fraction  -  is  contained  in  1, 

PROCESS  ^ 

c  1 

c      a      d      ad      ^  times ;  and  -  is,  therefore,  contained  in  1,  -  part 

b      d      b      c       be        t  J  ^-  d  .. 

of  d  times,  or  -  times, 
c 

Since  -  is  contained  -  times  in  1,  it  will  be  contained  -  times  -,  or  — 
d  c  '     h  c         be 

times  in  -• 

^  OR 

To  find  the  quotient  of  ^-r--. 
b     d 

§  102,  X  X  -  =  ^• 

d     b 

Multiplying  each  member  of  this  equation  by  -, 

c 
c     d     a     d 

XX-X-  —  -X-- 

d     c      b     c 

That  is,  x  =  ^x^'  (2) 

be 

Hence,  eq.  (1)  and  (2),  Ax.  1,  ^  ^  -  =  -  x  -• 
b     d     b     e 


FRACTIONS  155 

EuLE.  —  Multiply  the  dividend  by  the  reciprocal  of  the  divisor. 

1.  Change  integral  and  mixed  expressions  to  the  fractional  form. 

2.  An  integer  may  be  expressed  as  a  fraction  by  writing  1  for  its  de- 
nominator. 

3.  When  possible,  use  cancellation. 

Divide : 

2.  1  by  —  4.    1  by 

3.  Ibyi  '-'^y^^s 


Write  the  reciprocal  of 


10. 


b  n  b  —y 

7.   ^.  9.    ^.  U.    4- 

p  om  ab 

Simplify : 

,„    5mn  ,  10  m^  „,      a?  +  x  —  2 


Qbx 

'    3aa^ 

12  a'b 

4ax 

25  ac 

■  24  c2 

Sabm 

7 

-i-a^x. 

5ab 

25  b^ 

3aV 

'  15  a' 

7af  . 

21  x'f     • 
14  a 

my  — 

f   .       f 

■    25ac  ■  24c2 

14.  ^-^^^abx.  23. 

7 

,^     5a&      25  &2 

15.  —-TT^  -s-  ^  ^    „•  24. 


ar^  —  5a;  +  4 
a* -6* 


a^- 

-  X  — 

6 

0^  + 

X  — 

20 

.   a' 

+  b' 

a'-2ab  +  b'     a'-  ab 

x^  +  f  .  a^-\-ocy  +  y' 
x^  —  y'  x  —  y 

a?+h^      gg  _  aft  4- 1/ 
a2_4&2  •      a -26 


7_^_^21^V.    '  25      m^  —  if       m^  +  m'^y  +  my^ 

my  —  y*  my^  +  y^ 


17.    ."'■^-y   ^ 2^ 26. 


m*a;  +  m*      m^a^  —  ma;* 


(m  +  yY     m'  —  if  m^x  —  ma^       im?v?  +  a^ 


20 


156  FRACTIONS 

x^-6x^+llx~6    .         0^-13  a; +  12 
x^  +  2a^~19x-20  '  a^  + lOa^  +  29a;  + 20* 

ar^-15a^  +  74a;-120  .  a^ -9a^ +  26x-24: 
3^-5x2-07  +  5        ■   a^-ex'  +  llx-e' 

a2  +  62_c2  +  2a6  .  g' -b'' +  (f -2ac 
a^^h^-(?  +  2bc  '  a?-h^  +  c'  +  2ac 

a^  -\-  x^  —  y^  -\- 2  ax  _^  a?  —  a?  +  y^  -\- 2  ay 
a^  —  x^  +  y^  —  2ay     a^  —  x^  —  y^  —  2ocy 


30. 


31. 


32. 


33. 


34.   Simplify  (3,-0. +  ^^^1  +  1 


Solution 


Simplify : 


(--7)^(1-^) 


_y^  —  xy  +  x"^  .  x^  +  y^ 


y 

X2y2 

_x^-xy  +  y^^^ 

a;V 

y           (• 

x-^y)( 

x^  —  xy  +  J/*) 

_    x^y 

x  +  y 

H^ 

4:X  - 

0^- 

-^} 

f.,^''  +  ''l'. 

'^^f. 

:-\-S-4 

1 

37.    (af-^^fx-'^ 


38. 


40. 


i  +  i  +  iUri+l+i 


y^    y*. 


y    y 


39.    fl-n-(l-^  +  -^ 
x^J      \         y       y\ 


2x 


2f 


^■\-f. 


^    1- 


1]L 

^  +  y. 


FRACTIONS  157 


COMPLEX  FRACTIONS 

183.   A  fraction  one  or  both  of  whose  terms  contains  a  fraction 
is  called  a  Complex  Fraction. 

It  is  simply  an  expression  of  unexecuted  division. 

Examples 

- 

1.    Simplify  the  expression  -• 

y 

PROCESS 


£_a_^x_a     y  _ay 
X     b      y      b      X      bx 

y 

Simplify  the  following  expressions : 

x  +  y  2  +  — 

ab  4  b 

—  «  +  - 


ab^                                     3 
3. *  6.    j.  9. 

1 

X 

X 

7.    Y"  1®' 

11.   Simplify  the  expression  ^ — 

-2  +  -  +  1 

y^    y 

Solution. — Multiplying  the  numerator  and  denominator  of  the  fraction 
by  p%  the  L.  C.  D.  of  the  fractional  parts  of  the  numerator  and  denominator, 

the  depression  becomes  ^  —  ^V  +y . 

x'^  +  xy  +y' 


h  + 

c 

a 

Sm 

m  - 



X 

X 

X  - 



m 

6- 

c 
2 

b 

2 

c 

ax 

2 

a' 
2 

-ax 

x  +  y 

aj 

+  y 

y 

X 

1 

1 

2^ 

X 

158  FRACTIONS 

Simplify  the  following : 

12.    -^^.  14.      ,      "^^         „•  16. 


1         1 

+ 


x^-irf 

xy 

ar' 

~xy-\-f 

1 

a  +  1 

1 

1 

x'  +  y 

2 

X 

22/ 

a; 

y 

a; 

13.  ^:  yf-       15.    ^+!  .         17. 1:1^. 


a;     2/  +  2!                              a  +  1  1  +  a 

a;-2+— ^  6a-l-- 

18.   ^1.  20.    -^ ^• 

x  +  2+-^  ^^-Zl 

x-2  3a 

14      4  a; -5      ^24 

,^     X     x^     a?  2                X 

19. ^.  21. 


1  +  ^V^„ 


3a;-9 


.'c     a;-  X 


^  +  1         a;  +  l      1  —  a; 
22.    —. h  H 


\-\-x     \  —  x     \-\-x 


X—\       V  —  \       z  —  1 


3  xyz  X  y 

23  ^  -^ 


?/z  +  zx  +  an/  1      1      1 

X      2/      ^ 

12  9  a^  +  (a  +  &)a;  +  a6 


x  +  y     X  —  y     ^x  —  y  a^  —  (a  -\-b)x  +  ab 

24.    — ^-^ f ^.  25.    ^   J^    ;„ 

—  8y  a.-^  —  ft'' 


2/^  —  9  a^  *  ^  —  (^ 


i+  1 


a      6  +  c 
26 


1 1_  •  ^      y'  +  d'-a? 

a     h  +c  2  he 


FRA  CTIONS  159 


x  +  y     a^  +  y^ 

x  —  y  _  a?  —  -if     oj*  +  xh^  +  2^ 
x  +  y     a^  +  f 

a  —  hh  —  c  ^      c 


^^        1  +  a&     1  +  6c 


29. 


^        (a-6)(6-c)     -1 
(1  +  db)  (1  +  6c)     a  "^ 

(g  +  yf  -z"      {x-y  +  zy 
f  4. 


184.   An  expression  of  the  form   is  called  a 

Continued  Fraction.  b-\ 


d  + 


/+ 


30.    Simplify 


1+^ 


^-l 


Solution 
1 


1  +  -^     1  + 


14-^  ^  +  i 

X  X 

1 


1  + 


x+1 
_     x  +  1 
~  x  +  1 +z 

_  a;  +  1  , 
~2x+i 

SuGGESTTON.  —  In  the  above  example,  the  part  first  simplified  is  the  last 

complex  part r,  which  is  reduced  to  a  simple  fraction. 

1  +  - 

X 

Every  continued  fraction  may  be  simplified  by  successively  reducing  its 
last  complex  part  to  a  simple  fraction. 


160  FRACTIONS 

Simplify  the  following : 
1 


31. 


32. 


SS. 


.+     1 


1+^  +  1 


3  —  a; 

a 


a  +  lH 


a  +  1-- 
a 


2- 


34. 

x-2 

X      ^             a; 

35. 
36. 

x~2 
1 

.a  +  - 
a 

1 1       " 

'  1+0+    2c 

2-^  1  +  1 

2  —  a;  c 


REVIEW   OF   FRACTIONS 
185.   Reduce  to  their  lowest  terms  : 

^        a^-f  a^  +  a;-3  a;^  +  a:^ -  22 a; -  40 

a^  +  Sx'  +  Sx  +  S  '   a^-7a^  +  2a;  +  40' 

2  ^-i^-x-2  a!«  +  10a^  +  7a;-18. 
'«»  +  3ar'  +  3a;  +  2*                      '   a^-Sx'-llx  +  lS 

3  a^  +  4a^  +  8a;  +  5  2a^  +  7a;''-9a!-9 

■    a^  +  3a;2  +  7a;  +  5'  '     2a^  + 9aj2  +  x-3  ' 

^     a^  +  3ar'  +  4a;  +  2  8g^  -  22a^  + 17a; -3 

•a;3_3aJ!_8a;-10*  '   6a^ -17  a^ -i-Ux-3 

Simplify : 

9      ^    +    y    -  y-^ 

■    2y-1^2y  +  l      l-Ay^ 

10.   — ^! (      g        +-^ l^l^Y 

4(l-a)2     V8(l-a)      8(a  +  l)      4(a  +  l)y 

11     2a;-l      3a;4-l      3a;-l      2a;  +  l 
x  —  1        x  +  1        x  —  2        x  +  2' 

12. ^^^ + ^— +  "" 


{a  —  6)  (a  —  c)      (6  —  c)  (6  —  a)      (c  —  a)  (c  —  6) 


22. 


13. 
14. 
15. 
16. 


FRA  CTIONS 

m-  4-  m  —  2"* 


161 


fl  +      2     \  /m^  +  m  -  2Y 
\^        m  —  ly\    m^  +  m    J 


1 

a  +  a' 


af     X 


a 


y 


+  1 


17.      1  + 


a  —  b      a  +  6     a^  +  6'^ 

a;" 


4  a6  \  a^  —  6' 


y. 


18. 


19. 


20. 


x+l+^  +  h^ix  +  1 
X     ctr^ 

a  +  &     a^  +  6^  /a  +  6 


X     x^J 


a  —  b     a^  —  b^J\a  —  b 


63 


x^  +  y' 


\x-y         J 


21.    (a^-/ 


z^ 


2yz)^ 


x-\-y-\-z 


1  -f  a        4ct  8  a 


1-al  +  a'l-a^     1  +  a 


1- 


l  +  g'       4  a'' 

a"      1  +  a^     1  + 


23. 


a^a^  .  abx'  _  aca^  _  &^  .  a^  _  a 
6d       (^d        cP       ccP      be      d 


ax     b 
c      d 


24.    (x'-3xy-2y'+    ^^^  \^(3x-6y-     ^^   Y 


25. 


26.    1 


m  —  3  ri\  A         4  n   \  ^  /^^  ,  o  _  ^^  ^'^A 

2a;  +  5ar      (a^+x 


2(x  +  iy 


2x- 


27.      1  + 


a  —  a;y  va;  +  a 

ALG.  — 11 


/3-3a;\ 


X' 

2  ay'  -\- 2  ax  —  a 
3c^  +  3ax  +  2a^ 


162  FRACTIONS 

Expand : 

Simplify : 

1  4_i  ±.—                                          6a  4 

34. ^.  38.  -^    -^. 


1     1\  a2  +  3a  +  2 

—  7>l  1^1 


35.    -^^ ,   .^    / ^—  39.    1- 


m^  —  n^  g^  +  5  g  +  4 

d  m^  —  w^  A         2  w 


.+ 


3g_    (g  +  1)''      (g  +  iy  ^^     m«  +  n' 


u,  1  '      -J 2m?i 

(g  + 1)*      (g  +  1)*  m^  —  mn  +  w^ 

37.    z 41.    


1- 


3 


4-    ' 


1  —  X  6  —  X 

'    \x  —  y     x  +  y)  '  \2(x  —  y)      2x+2yj 

(i+-+-+-Mi--+-> 

\ar     x     a      ay  \oir     x     aj 


'   Qt?f-a?y'^  2        4       a;?/  + 1 

xy     x^y^ 


SIMPLE    EQUATIONS 


ONE  UNKNOWN   NUMBER 

186.  An  Equation  has  been  defined,  §  2,  as  an  expression  of 
equality  between  two  numbers  or  quantities. 

187.  An  equation  all  of  whose  known  numbers  are  expressed 
by  figures  Us  called  a  Numerical  Equation. 

188.  An  equation  one  or  more  of  whose  known  numbers  is 
expressed  by  letters  is  called  a  Literal  Equation. 

189.  -  An  equation  that  does  not  involve  an  unknown  number 

in  any  denominator  is  called  an  Integral  Equation. 

2  X 
sc  +  5  =  8  and h  5  =  8  are  integral  equations.     The  second  equation 

is  integral ;  for  though  it  contains  a  fraction,  the  unknown  number  x  does 
not  appear  in  the  denominator. 

190.  An  equation  that  involves  an  unknown  number  in  any 
denominator  is  called  a  Fractional  Equation. 

8  Ix 

a;  +  5  =  -  and  =  7  are  fractional  equations. 

x  x  —  \ 

191.  An  equation  whose  members  are  identical,  or  such  that 
they  may  be  reduced  to  the  same  form,  is  called  an  Identical 
Equation,  or  an  Identity. 

a  +  b=a  +  b  ;  a^  —  b^  =  (a  -i-  b)(a  —  b)  are  identical  equations. 

An  equation  whose  members  are  numerical  is  evidently  an 
identical  equation. 

10  =  6  +  4;  8x2  =  6  + 12  —  2  are  identical  equations. 

163 


164  SIMPLE  EQUATIONS 

A  literal  equation  that  is  true  for  all  values  of  the  letters  in- 
volved is  an  identical  equation,  or  an  identity. 

(x  +  yY  =  x^  +  2xy  +  y^  is  au  identity,  because  it  is  true  for  all  values  of 
X  and  y. 

192.  An  equation  that  is  true  for  only  certain  values  of  its 
letters  is  called  an  Equation  of  Condition. 

An  equation  of  condition  is  usually  termed  simply  an  Equation. 

X  4-  4  =  10  is  an  equation  of  condition,  because  it  is  true  only  when  the 
value  of  a;  is  6.  a;^  _  9  jg  ^q  equation  of  condition,  because  it  is  true  only 
when  the  value  of  a;  is  +  3  or  —  3. 

193.  When  an  equation  is  reduced  to  an  identity  by  the  substi- 
tution of  certain  numbers  for  the  unknown  numbers,  the  equation 
is  said  to  be  satisfied. 

When  «  =  2,  the  equation  3  x  +  4  =  10  becomes  6  +  4  =  10,  an  identity  ; 
consequently,  the  equation  is  satisfied. 

Any  number  that  satisfies  an  equation  is  called  a  Root  of  the 
equation. 

2  is  a  root  of  the  equation  3  x  +  4  =  10. 

Finding  the  roots  of  an  equation  is  called  solving  the  equation. 

194.  An  integral  equation  that  involves  only  the  first  power  of 
one  unknown  number  in  any  term  when  the  similar  terms  have 
been  united  is  called  a  Simple  Equation,  or  an  Equation  of  the  First 
Degree. 

3 X  +  4  =  10  and  x  +  22/  =  z  +  8  are  simple  equations. 

195.  Two  equations  that  have  the  same  roots,  each  equation 
having  all  the  roots  of  the  other,  are  called  Equivalent  Equations. 

By  the  axioms  in  §  74,  if  the  members  of  an  equation  are 
equally  increased  or  diminished  or  are  multiplied  or  divided  by 
the  same  or  equal  numbers,  the  resulting  numbers  are  equal  and, 
§  186,  form  an  equation.  But  it  does  not  necessarily  follow  that 
the  equation  so  formed  is  equivalent  to  the  given  equation. 

For  example,  if  both  members  of  the  equation  x  +  2  =  5,  whose  only  root 
is  X  =  3,  are  multiplied  by  x  —  1,  the  resulting  numbers  (x  +  2)(x  —  1)  and 
5(x  —  1)  are  equal  and  form  an  equation  (x  +  2)  (x  —  1)  =  5(x  —  1)  ;  but  this 
equation  is  not  equivalent  to  the  given  equation,  since  it  is  satisfied  by  x  =  1 
as  well  as  by  a;  =  3. 


SIMPLE  EQUATIONS  165 

196.  Principles.  —  ^-  If  the  same  expression  is  added  to  or 
subtracted  from  both  members  of  an  equation,  the  resulting  equation 
is  equivalent  to  the  given  equation. 

2.  If  both  members  of  an  equation  are  multiplied  or  divided  by 
the  same  Tcnown  number,  except  zero,  the  resulting  equation  is  equiva- 
lent to  the  given  equation. 

3.  If  both  members  of  an  integral  equation  are  midtiplied  by  the 
same  unknown  integral  expression,  the  resulting  equation  has  all  the 
roots  of  the  given  equation  and  also  the  roots  of  the  equation  formed 
by  placing  the  multiplier  equal  to  zero. 

It  follows  from  Prin.  3  that  if  the  same  unknown  factor  is  removed 
from  both  members  of  an  equation,  the  resulting  equation  has  all  the  roots 
of  the  given  equation  except  those  obtained  from  the  equation  formed  by 
placing  the  factor  removed  equal  to  zero. 

Principle  1  may  be  established  as  follows : 

Let  A  =  B  (1) 

be  any  equation  and  C  any  expression  to  be  added  or  subtracted. 

It  is  to  be  proved  that  A±  C=  B  ±C  .  (2) 

is  equivalent  to  (1),  the  given  equation. 

All  the  values  of  the  unknown  number  or  numbers  that  satisfy  (1),  that 
is,  make  A  identical  with  B,  make  A  +  C  identical  with  B  +  C  and  A  —  0 
identical  with  B  —  C,  that  is,  satisfy  (2).     Hence,  (2)  has  all  the  roots  of  (1). 

For  the  same  reason     A±CT  C=  B±  CT  C  (3) 

has  all  the  roots  of  (2).  But  by  the  Associative  Law  for  addition  (3)  may  be 
written  A  +  0  =  B  +  0,  or  A  =  B.     Hence,  (1)  has  all  the  roots  of  (2). 

Since  (2)  has  all  the  roots  of  (1)  and  (1)  has  all  the  roots  of  (2),  (2)  is 
equivalent  to  (1),  the  given  equation. 

Principles  2  and  3  may  be  established  as  follows : 

Let  A  =  B  (1) 

and  MA  =  MB.  (2) 

From  (1),  by  Prin.  1,  A-B  =  0.  (8) 

From  (2),  by  Prin.  1,    M(A  -  S)=  0.  (4) 

Since  the  first  member  of  (4)  can  reduce  to  zero  only  when  one  or  both  of 
its  factors  become  0,  (4)  is  satisfied  by  those  values  of  the  unknown  number 
that  make  A  —  B  =  0,  that  is,  by  the  roots  of  (3),  or,  Prin.  1,  of  (1);  and 
also  by  those  values  of  the  unknown  number  that  make  M=0,  that  is,  by 
the  roots  of  M  =  0,  but  by  no  other  values. 

If  M  is  any  known  number,  not  zero,  M  cannot  be  placed  equal  to  zero 
and  then  (4),  or  (2),  is  equivalent  to  (3),  or  to  (1). 


166  SIMPLE   EQUATIONS 

If  Jf  is  an  unkuown  expression,  (4),  or  (2),  lias  the  roots  of  itf  =  0  in  addi- 
tion to  the  roots  of  ^  —  ^  =  0,  or  of  (1). 

197.  By  §  196,  Prin.  1  and  2,  every  simple  equation  involving 
one  unknown  number  may  be  reduced  to  an  equivalent  equation 
having  the  form  x  =  a,  a  being  a  fixed  known  number.     Hence, 

Every  simple  equation  involving  one  unknown  number  has  one 
root  and  only  one  ;  also,  by  §  196,  Prin.  3, 

The  equation  (x  —  a)  {x  —  h)  {x  —  c)  •••  (x  —  r)  =  0  is  equivalent 
to  the  simple  equations  x  —  a  =  0,  x  —  b=:0,  x  —  c=0,  '■•  x  —  r  =  0, 
and  has  as  many  roots  as  factors  of  the  first  degree  involving  x. 

CLEARING   EQUATIONS    OF   FRACTIONS 

198.  1.    If  one  third  of  a  number  is  10,  what  is  the  number  ? 

2.  li  \x  =  l,  what  is  the  value  of  aj?  li  \x  =  5,  what  is  the 
value  of  a;  ?     If  i  a;  =  6,  what  is  the  value  of  a;  ? 

3.  If  I  a;  =  6,  what  is  the  value  of  2  a;?  If  ^a;  =  10,  what  is 
the  value  of  5  a;  ? 

4.  If  -  =  5,  what  is  the  form  of  the  resulting  equation  when 

o 

both  members  are  multiplied  by  3  ?  by  6  ?  by  9  ?  by  any  multiple 
of  the  denominator  ? 

199.  The  process  of  changing  an  equation  containing  fractions 
to  an  equation  without  fractions  is  called  Clearing  the  Equation  of 
Fractions. 

200.  Principles.  —  1.  An  equation  may  he  cleared  of  fractions 
by  multiplying  both  members  by  some  common  multiple  of  the  denomi- 
nators of  the  fractions.     (Ax.  4.) 

2.  If  both  members  of  a  fractional  equation  are  midtiplied  by  the 
eocpression  of  loioest  degree  required  to  clear  the  equation  of  fractions, 
the  resulting  equation  is  equivalent  to  the  given  equation. 

Principle  2  may  be  established  as  follows : 

By  §  196,  Prin.  1,  all  the  terms  of  the  second  member  may  be  transposed 
to  the  first  member.  Hence,  uniting  the  terms  of  the  first  member  into  one 
and  reducing  this  to  its  lowest  terms,  any  fractional  equation  may  be  reduced 
to  an  equivalent  equation  of  the  form 

4  =  0  0) 


SIMPLE   EQUATIONS  167 

in  which  A  is  prime  to  J5,  and  B  is  the  expression  of  lowest  degree  required 
to  clear  the  equation  of  fractions. 

Since  A  and  B  have  no  common  factors,  A  and  B  cannot  reduce  to  zero 
at  the  same  time  for  any  value  of  the  unknown  number.  Hence,  eq.  (1) 
is  satisfied  by  every  value  of  the  unknown  number  that  makes  ^  =  0,  and  by 
no  other  values ;  that  is,  the  equation  ^  =  0,  obtained  by  multiplying  both 
members  of  the  given  equation  by  the  expression  of  lowest  degree  required 
to  clear  it  of  fractions,  is  equivalent  to  the  given  equation. 

Examples 

1.    Given  '- =6 ,  to  find  the  value  of  x. 

4  3 

PROCESS 

a;  — 4_^      X 


Clearing  of  fractions,  3a;  — 12  =  72  —  4a; 

.-.  a;  =  12 

Explanation.  —  Since  the  first  fraction  will  become  an  integer  if  the 
members  of  the  equation  are  multiplied  by  4  or  any  multiple  of  4,  and  since 
the  second  fraction  will  become  an  integer  if  the  members  of  the  equation  are 
multiplied  by  3  or  any  multiple  of  3,  the  equation  may  be  cleared  of  fractions 
in  a  single  operation  by  multiplying  its  members  by  some  common  multiple 
of  4  and  3  (§  196,  Prin.  2,  §  200,  Prin.  1). 

Since  the  terms  derived  from  the  numerators  will  be  least  or  lowest  when 
the  multiplier  is  the  least  or  lowest  common  multiple  of  the  denominators, 
the  members  of  the  equation  should  be  multiplied  by  the  least  common  mul- 
tiple of  4  and  3,  which  is  12. 

The  resulting  equation  is  3  a;  —  12  =  72  —  4  a; ;   .-.  a;  =  12. 

X  —  1 
2.    Find  the  value  of  x  in  '■ 


x-2     2 
3        3 

X-3 
4 

2      x-S 

3         4 

Solution.  

2  3 

Clearing  of  fractions,  6(x  -  1)  -  4(x  -  2)  =  8  -  3(a;  -  3). 

6a;-G-4a;  +  8  =  8-3«  +  9. 

.-.  X  =  3. 


168  SIMPLE   EQUATIONS 

Rule.  —  Multiply  both  members  of  the  equation  by  the  least  or 
lowest  common  multiple  of  the  denominators. 

1.  The  multiplier  of  lowest  degree  required  to  clear  an  equation  of  frac- 
tions so  that  the  resulting  equation  is  equivalent  to  the  given  equation  is 
usually  the  L.  C.  D.  But  if  fractions  having  like  denominators  have  not 
been  united  and  every  fraction  reduced  to  its  lowest  terms,  the  multiplier 
required  may  be  of  lower  degree  than  the  L.  C.  D. 

Thus,  given  ^^  =  ^J^±^  +  ^^±^. 

%  —  'I       X  —  2  3 

Uniting  terms,  2x2-.Sa;-2^  or  2  a;  +  1  =  'i*-±-§. 
X  —  2  3 

Multiplying  by  3,  6  a;  +  3  =  5  x  +  8. 

.*.  X  =  5. 

Had  the  given  equation  been  cleared  of  fractions  by  multiplying  by 
3(x  —  2)  instead  of  by  3,  the  resulting  equation  simplified,  which  is 
aj'^  —  7  X  +  10  =  0,  or  (x  —  5)  (x  —  2)  =  0,  would  have  been  satisfied  both  by 
X  =  5  and  by  x  =  2.  The  given  equation  is  satisfied  by  x  ==  5  but  not  by 
X  =  2.  Hence,  the  latter  value  is  not  a  root  of  the  given  equation,  but  has 
been  introduced  by  using  (x  —  2)  times  the  necessary  multiplier,  3.  Boots 
so  introduced  may  be  discovered  by  verification  and  rejected. 

2.  If  a  fraction  has  the  minus  sign  before  it,  the  signs  of  all  the  terms 
of  the  numerator  must  be  changed  when  the  denominator  is  removed. 

Find  the  value  of  x,  and  verify  the  result : 

3.  2x  +  -^~.  7.    ^  +  ^  =  1^. 

3       3  2      6       3 

4.  -  +  10  =  13.  8.    7|-  — =-. 

5.  -  +  2x  =  26.  9.    ^  +  ^  =  24. 
6  5      7 


6.    3a;--  =  14.  10. 


3 

11. 

XX     X     3x      o  X  _j 
2     3     4  "^10       12  ~    ■ 

25  a;      5a;2«     5a;_o 
18        9        3        6  ~    ■ 

12. 

2  X     5  _  a; 
6~4* 


2 X     ^' _x  ,  4^ _ ^ 

'       Y~4:~2^T~U 


SIMPLE   EQUATIONS  1(39 


2x      7a;      5a;       a;  __  4 
T       8       18      24  ~  9* 

T^16      2"    16      8' 

,„     15  a;  ,  5  a;      11a;  ,  19  a;      « 

16. =  ^« 

7         6         3         14 

,„     2a;  ,  5a;      4x  ,  a;      1 

17. —  =  — 

15      25       9       6     9 

3a;      7x^11  a;      8a;3 
4       12        36        9       2* 

19     3a;     JL^_3a;     X     4^^ 
^^'   T^20"10^5+15 


20.    I^±i-2a;  +  ^^-^±^  =  l. 
5  7 


21. 

a;  —  la;  —  2x  —  3      5  a;  —  1 
2             3             4              6 

22. 

x  +  1      a;  +  4a;  +  3_-|^g 
3            5*4 

23. 

7a;  +  2      12-a;a;  +  2_^ 
6               4              2           * 

24. 

a;-3a;  +  r>      a;  +  2      ^ 
7             3             6 

25. 

3  a;  -  5      7  a;  -  13  _  3      a;  +  3 
4                6                      2 

26. 

2a;-5      3a;-2_-,^      a;  +  2_ 
5               7                    6 

27. 

l-2a;     7-2a;     ll-2a; 
3                4*6 

7 
12' 

28. 

a;  +  4,2-2a;      a;+l      qi 
3      '       6             2        ^* 

170  SIMPLE   EQUATIONS 


29     9a;  +  5      8a;-7  ^  36  a;  +  15      10^ 
14      "^  6a; +  2  56  14* 

Suggestion.  —  The  equation  may  be  written 

§104,3,  9»     j_     8a;-7^36^     15     41^ 

14      14     6  a; +  2       66       56     56 

Simplify  as  much  as  possible  before  clearing  of  fractions. 
„_     3x-2  ,  8a7-21      6a;-22 


31. 


32. 


33. 


34. 


35. 


2  a; -5  5  10 

4a;  +  3^8a;  +  10      7 a; -29 
9  18  5  a;  - 12* 

6a;  +  l       2a;-4  ^2a;-l 
15          7a;-13~       5 

10  a;  +  17      5  a;  -  2  ^  12  a;  -  1 
18  9       ~lla;-8' 

6a;  +  3       3a;-l  ^2a;-9 
15  5x-25~      5 

2a;  +  l  8       ^2a;-l 

2a;-l      4a;2-l~2a;  +  l' 


36.    Solve  the  equation  ^Jzl  +  ?J:z:§  =  £lZ^  +  ^JH^. 
^  a;-2a;-7      a;-6a;-3 

Solution.  —  It  will  be  observed  that  if  the  fractions  in  each  member  were 
connected  by  the  sign  — ,  and  the  members  were  simplified,  the  numerators 
of  the  resulting  fractions  would  be  simple.  The  fractions  can  be  made  to 
meet  this  condition  by  transposing  one  fraction  in  each  member. 

Consequently,  it  is  sometimes  expedient  to  defer  clearing  of  fractions. 

Transposing,  *~^     x-2     x-b     x- 


Uniting  terms, 


X  - 

-1 

-2 

X  — 

2 

X  - 

X  — 

3 

— 

1 

x-6     x-1 

-1 
a;2  _  5  X  +  6     «^  -  13  a;  +  42* 


Since  the  fractions  are  equal  and  their  numerators  are  equal,  their  denomi- 
nators must  be  equal. 

x2  _  5  X  +  6  =  a;2  _  13  X  +  42. 

.-.  X  =  4i. 


39. 


40. 


SIMPLE  EQUATIONS  171 

X—Z       X  —  Q       03—6       x  — 4 

„Q     x  —  ^x  —  1      x  —  Q     X  —  ^ 

on. -\ -  —  — —  H -• 

iK  —  4      x  —  Q      X  —  7     X  —  5 

x  +  2  _  x  +  3  _  x  +  5  _  x-\-6 
x  +  1      x-\-2~  x  +  4:     x  +  5' 

x-\-l      x  +  6  _x  +  2     x  +  5 
x  +  2      x-\-7~x  +  3     x  +  G 

.^     a;  — 5_a;  — 10_ir  — 4_a;  — 9 

'x  +  5      a;  +  10~ic  +  4     x  +  d 

I^_2     5a;--     ^-2 

-  A^ r^=V-i- 

43.  ^nil  +  '^^  +  ^.zie.s^o. 

oj  —  2      x  — 3      X  —  4: 

44.    ■ = \-  2x. 

x-1       x  +  1      x^-l 

45.    —- T  =  —o — T  — 2  a; 

a;  +  l       a;  —  1      ar  —  1 

46.  i-^  =  Ii^ +  ?('?- A 

2  3  2l^a;       y 

A-16     A  +  6 
47     bx  ox         _4^ 

24  60~^y 

48.   |(2-a;)-|(3-2a;)=^±12. 

2x(\-^^      3x{\-^ 

V        x)  \        xj      a; -4 

*'•    3 + 4 -J- 

50.    l.-2(^-3)=4-|(^^l). 

61     (2a;  +  l)'      (4a;-iy^l5     3(4  a; +  1) 
6  20  8  40       ' 


172  SIMPLE   EQUATIONS 

52.    l.^±l  +  2  =  l  +  '^-l  ^ 


2a;-l  2a;  +  l      l-a:^ 

174_?      i+l§     21_^      100     5 

2x 


54.    K^-4)      4a;-16^3 

2 

10 


+  5 


G  5  f 


2  + 

55.    1+ T  = jr'  56.    =  — T^- 

1  4-1       1+^  x_^  x  +  5 

X  X  x     3 


3      3a;-l 


1.    Solve  the  equation 


Literal  Equations 

X  —  W      X  —  a^ 


a  b 

Solution 
a;  -  62     X  -  a2 


a  b 

Clearing  of  fractions,  bx  —  b^  =  ax  —  a". 

Transposing,  etc.,  ax  —  bx  =  a^  —  b^. 

(a  —  b)x  =  a^  —  b^. 
Dividing  by  (a  —  6) ,  x  —  a^  +  ab  +  b^. 

Solve  the  following  equations : 

2     ^  —  ^  I  ^^^ _  1 ,  g     a;  —  2a&      l_a;  —  3c 

na;         ca;     c  cx  x         abx 

^  _a6__7  __49^  _     a?  —  a|2a;_q      6  b 

X      ab     abx  ha  a 

4     .^_2a!=i_?_^.  8     «'  [   b''  ^a  +  h     3(a  +  6) 

a6^      ft^a;  a^a;  bx     ax        ah  x 

K     ^     a;4-26_a      ^^  _     a^  +  6^     a  —  h  _b 

b  a      ""6        '  '      2bx   ~  2bx'~  x 


SIMPLE  EQUATIONS  173 

iA         I  h{x  +  g)  _  2  a5  a;  —  2a  ,  x _a-  -\-h^ 

10.    ft  H  — "  11.  f- —  — — • 

X  —  a       X—  a  a  o         ao 

12.  Q X  +  l^(l -^=  a{x  -  a). 

13.  6(2a;-9c-14&)=c(c-a;). 

14.  a(x-a-26)+6(x- &)+c(x  + c)=0. 

15.  (ft  —  x){x  —  5)  +  (ft  +  a;)(a;  —  6)  =  (ft  —  6)-. 
a  —  ft  +  c      h  —  a  -\-  c 


16. 


a;  +  ft  x  —  a 


17.    ^^ + -^ =  0. 

ft(6  —  x)      b{c  —  x)      a{c  —  x) 

1-A  0  0 

ft  —  1  X-  —  a- 


a  -  1  x-1  (ft  -  l)(.r  -  1) 

a  +  x  2x  x^(x  —  ft)  _  1^ 

ft  ft  +  a;  ft(ft^  —  a:^)     3 

20     ^  +  «  I  -^  +  c  a;  +  ?)  _  ft      6      c       j^^ 

6  ft             c         b      c     a 


21. 


22. 


a:^  —  fta?  —  6a;  -f-  ft5      a^  —  2  6.t  +  2  />^         c^ 


a;  —  ft  X  —  b  X  —  c 

1  2  m?j  7n  X  —  lb 


m  -f  w      (m  +  7(-)''*      (m  +  ny      (m  +  ^0^ 


23. + =a?  +  b''  +  c'  +  2db. 

a  -{■  b  +  c     a  -\-b  —  c 

a;  4-26      a;  +  3  6  ^   a;  4-  6    ,  a;  +  26 
X  —  b        X  +  b       X  —  Sb     a;4-56 

-_     2a;  +  3ft,3a;4-7ft         2a       ,m 

25.     ' 1 —: =  ' 1-5- 

a;4-a  a;  +  2a       a;  +  4a 

„-     a;  +  7a.    X  —  a       x  +  7  a       x  —  a 
^D. 1 —  = 1- 


a;-f6ft      x  —  3ft       a;  +  ft       ic  +  2ft 
27.    (ft  —  6)(a!  —  c)  —  (6  —  c)(x  —  a)  =  (c  —  a)(x  —  6), 


174  SIMPLE  EQUATIONS 

Problems 

Directions  for  Solving.  —  Represent  one  of  the  unknown 
numbers  by  x,  and  from  the  conditions  of  the  problem  find  an 
expression  for  each  of  the  other  unknown  numbers. 

Find  from  the  problem  two  expressions  that  are  equal,  and  write 
them  as  an  equation. 
Solve  the  equation. 

1.  A  man  bought  a  farm,  a  house,  and  a  barn  for  $12,600. 
If  the  house  cost  twice  as  much  as  the  barn,  and  the  farm  twice 
as  much  as  the  house  and  barn  together,  how  much  did  each 
cost? 

Suggestion. — There  are  three  unknown  quantities  —  the  cost  of  the 
house,  the  cost  of  the  barn,  and  the  cost  of  the  farm.  It  is  evident  that,  if 
the  cost  of  the  barn  were  known,  the  cost  of  the  house  could  be  found  from 
it,  and  from  both  the  cost  of  the  farm. 

Accordingly,  represent  the  cost  of  the  barn  as  x  dollars,  and  express  the 
cost  of  the  house  and  the  cost  of  the  farm  in  terms  of  x.  Discover  two 
expressions  for  the  total  cost,  and  equate  them. 

2.  If  A  is  twice  as  old  as  B,  and  B  twice  as  old  as  C,  how 
old  is  each, if  the  sum  of  their  ages  is  140  years  ? 

3.  Mr.  Henry  bought  three  building  lots  for  $36,000.  If  the 
third  cost  twice  as  much  as  the  second,  and  the  second  3  times 
as  much  as  the  first,  what  was  the  cost  of  each  ? 

4.  A  man  left  $63,000  to  his  wife,  two  sons,  and  a  daughter. 
If  each  son  received  twice  as  much  as  the  daughter,  and  half  as 
much  as  the  wife,  what  was  the  share  of  each  ? 

5.  A  man  left  ^  of  his  property  to  his  wife,  \  to  his  son,  \  to 
his  daughter,  and  the  rest,  which  was  $2000,  to  a  hospital.  What 
was  the  value  of  his  property  ? 

6.  A  owed  B,  C,  and  D  $27,000  in  all.  If  he  owed  B  4  times 
as  much  as  C,  and  D  f  as  much  as  B,  what  sum  did  he  owe 
each  ? 

7.  A  person  spends  \  of  his  annual  income  for  board,  \  for 
clothes,  and  $260  for  other  expenses.  If  he  saves  ^  of  his 
income,  what  is  his  income  ? 


SIMPLE   EQUATIONS  175 

8.  A  table  and  a  chair  cost  $  11.  The  table  and  a  picture 
cost  $  14.  If  the  chair  and  the  picture  together  cost  3  times  as 
much  as  the  table,  what  was  the  cost  of  each  article  ? 

9.  A  and  B  together  had  $  50,  and  A  and  C  had  f  60.  After 
each  had  spent  $5,  A  had  \  as  much  as  B  and  C  together. 
How  much  had  each  at  first? 

10.  A  gave  his  age  as  follows  :  "  f  of  my  age  less  ^  of  what  it 
will  be  a  year  hence  is  equal  to  \  of  my  age  5  years  ago."  What 
was  his  age  ? 

11.  James  is  5  years  older  than  his  sister,  and  5  years  hence 
he  will  be  3^  times  as  old  as  his  sister  was  5  years  ago.  What  is 
the  age  of  each  ? 

12.  A's  daily  wages  are  f  of  B's,  and  C's  are  f  of  A's.  If  A 
and  C  together  earn  25  cents  more  a  day  than  B,  what  are  the 
daily  wages  of  each  ? 

13.  A  man  paid  a  debt  of  $8.00  with  an  equal  number  of 
6,  10,  and  25-cent  pieces.     How  many  of  each  were  there  ? 

14.  A  man  bought  equal  quantities  of  white  and  brown  sugar, 
paying  6^  cents  a  pound  for  the  former  and  5  cents  a  pound  for 
the  latter.  How  many  pounds  of  each  did  he  buy,  if  the  whole 
quantity  cost  him  $  1.80  ? 

15.  A  field  is  twice  as  long  as  it  is  wide.  By  increasing  its 
length  20  rods  and  its  width  30  rods,  the  area  will  be  increased 
2200  square  rods.     What  are  its  dimensions  ? 

16.  Three  fifths  of  a  certain  number  exceeds  ^  of  it  by  7. 
What  is  the  number? 

17.  One  third  of  a  number  added  to  3  times  the  number  is 
equal  to  50.     What  is  the  number  ? 

18.  After  spending  f  of  my  income  and  $300  more,  I  had 
\  of  it  left.     What  was  my  income? 

19.  The  sum  of  ^  of  a  number,  \  of  it,  and  |  of  it  is  4  more 
than  f  of  the  number.     What  is  the  number  ? 

20.  The  sum  of  a  number,  its  half,  its  third,  and  its  fourth, 
and  16  is  66.     What  is  the  number  ? 


176  SIMPLE   EQUATIONS 

21.  A  man  deposited  f  of  his  month's  wages  in  a  bank,  and 
paid  out  ^  of  the  remainder  for  groceries  and  $  9  for  dry  goods. 
If  he  had  $  3  left,  how  much  money  had  he  at  first  ? 

22.  Divide  500  into  two  parts,  such  that  the  greater  decreased 
by  \  of  the  smaller  is  5  times  as  much  as  the  smaller  decreased 
by  ^  of  the  larger. 

23.  Divide  61  into  two  parts,  such  that,  if  the  greater  is 
divided  by  the  less,  the  quotient  will  be  3  and  the  remainder  1. 

24.  Divide  111  into  two  parts,  such  that  the  first  diminished 
by  3  is  equal  to  the  second  divided  by  3. 

25.  Divide  40  into  two  parts,  such  that  the  first  divided  by  3 
is  equal  to  the  second  divided  by  5. 

26.  Divide  40  into  two  parts,  such  that  the  first  is  10  greater 
than  twice  the  second. 

27.  Divide  44  into  two  parts,  such  that,  if  the  greater  is  divided 
by  5  and  the  less  by  7,  the  difference  of  the  quotients  will  be  4. 

28.  Divide  54  into  two  parts,  such  that  ^  of  the  first  part  is 
equal  to  \  of  the  second. 

Solution 
Let  a;  =  J  of  the  first  part  or  ^  of  the  second  part. 

Then,  4  x  =  the  first  part, 

and  6  X  =  the  second  part ; 

4a;  +  5x  =  54; 
whence,  x  =  6, 

4  X  =  24,  the  first  part, 
and  5  X  =  30,  the  second  part. 

29.  Divide  40  into  three  parts,  such  that  ^  of  the  first,  ^  of  the 
second,  and  \  of  the  third  are  equal. 

30.  Find  three  parts  of  60,  such  that  the  first  divided  by  5, 
the  second  multiplied  by  2,  and  the  third  increased  by  5  are 
equal. 

31.  Divide  72  into  four  parts,  such  that  the  first  divided  by  2, 
the  second  diminished  by  2,  the  third  multiplied  by  2,  and  the 
fourth  increased  by  2  are  equal. 


SIMPLE  EQUATIONS  177 

32.  A  can  do  a  piece  of  work  in  8  days.  If  B  can  do  it  in  10 
days,  in  how  many  days  can  both  working  together  do  it  ? 

Solution 
Let  X  =  the  required  number  of  days. 

Then,  -  =  the  part  of  the  work  both  can  do  in  1  day, 

X 

I  =  the  part  of  the  work  A  can  do  in  1  day, 
^  =  the  part  of  the  work  B  can  do  in  1  day  ; 

X     8      10      40' 
Solving,         X  =  ^,  or  4|,  the  required  number  of  days. 

33.  A  can  do  a  piece  of  work  in  10  days,  and  B  can  do  it 
in  15  days.     How  long  will  it  take  both  to  do  it  ? 

34.  Three  pipes  empty  into  a  cistern.  One  can  fill  the  cistern 
in  5  hours,  another  in  6  hours,  and  the  third  in  10  hours.  How 
long  will  it  take  the  three  pipes  together  to  fill  it  ? 

35.  A  can  do  a  piece  of  work  in  10  days,  B  can  do  it  in  12 
days,  and  C  can  do  it  in  8  days.  In  how  many  days  can  all 
together  do  it? 

36.  A  can  pave  a  walk  in  6  days,  and  B  can  do  it  in  8  days. 
How  long  will  it  take  A  to  finish  the  job  after  both  have  worked 
3  days  ? 

37.  A  can  build  a  wall  in  15  days,  but  with  the  aid  of  B  and 
C,  the  wall  can  be  built  in  6  days.  If  B  does  as  much  work 
in  1  day  as  C  does  in  2  days,  in  how  many  days  can  B  and  C 
separately  build  the  wall  ? 

38.  A  and  B  can  dig  a  ditch  in  10  days,  B  and  C  can  dig  it  in 
6  days,  and  A  and  C  in  7^  days.  In  what  time  can  each  man  do 
the  work  ? 

Suggestion.  —  Since  A  and  B  can  dig  ^  of  the  ditcli  in  1  day,  B  and  C 
I  of  it  in  1  day,  and  A  and  C  ^  of  it  in  1  day,  i^  +  i  +  ^  is  twice  the  part 
they  can  all  dig  in  1  day. 

39.  A  and  B  can  load  a  car  in  2y\  hours,  B  and  C  in  3f  hours, 
and  A  and  C  in  3^  hours.  How  long  will  it  take  each  alone  to 
load  it  ? 

ALG.  — 12 


178  SIMPLE  EQUATIONS 

40.  A  boy  bought  some  oranges  at  the  rate  of  30  cents  a  dozen. 
He  sold  f  of  them  for  4  cents  each,  and  the  rest  for  3  cents  each. 
If  he  gained  90  cents,  how  many  oranges  did  he  buy  ? 

41.  Find  a  fraction  whose  value  is  ^  and  whose  denominator 
is  15  greater  than  its  numerator. 

42.  Find  a  fraction  whose  value  is  f  and  whose  numerator  is 
3  greater  than  half  of  its  denominator. 

43.  The  numerator  of  a  certain  fraction  is  8  less  than  the  de- 
nominator; and  if  each  term  of  the  fraction  is  decreased  by  5, 
the  value  of  the  fraction  becomes  \.    What  is  the  fraction  ? 

44.  The  units'  digit  of  a  number  expressed  by  two  digits  ex- 
ceeds the  tens'  digit  by  5.  If  the  number  increased  by  63  is 
divided  by  the  sum  of  its  digits,  the  quotient  is  10.     What  is  the 

number  ?  * 

Solution 

Let  z  =  the  digit  in  tens'  place. 

Then  a;  4-  6  =  the  digit  in  units'  place, 

and  10  re  +  (a;  +  5),  or  11  a;  -I-  5  =  the  number  ; 

2a;  +  5 
whence,  x  =  2, 

and  a;  +  6  =  7. 

Therefore,  the  number  is  27. 

45.  The  tens'  digit  of  a  number  expressed  by  two  digits  is 
3  times  the  units'  digit.  If  the  number  diminished  by  33  is 
divided  by  the  difference  of  the  digits,  the  quotient  is  10.  What 
is  the  number  ? 

46.  The  tens'  digit  of  a  number  expressed  by  two  digits  is  ^ 
of  the  units'  digit.  If  the  number  increased  by  27  is  divided  by 
the  sum  of  its  digits,  the  quotient  is  6^.     What  is  the  number  ? 

47.  In  a  purse  containing  $1.45  there  are  ^  as  many  quarters 
as  5-cent  pieces  and  f  as  many  dimes  as  5-cent  pieces.  How 
many  pieces  are  there  of  each  kind  ? 

48.  A  woman  spent  $10  more  than  f  of  her  money;  then 
$10  more  than  |  of  the  remainder.  If  she  had  $2  left,  how 
much  money  had  she  at  first  ? 


SIMPLM  EQUATIONS  179 

49.  A  man  spent  $  1  less  than  ^  of  his  money  and  had  left 
$  1  less  than  f  of  it.     How  much  money  had  he  at  first  ? 

50.  A  girl  found  that  she  could  buy  12  apples  with  her  money 
and  have  5  cents  left,  or  10  oranges  and  have  6  cents  left,  or  6 
apples  and  6  oranges  and  have  2  cents  left.  How  much  money 
had  she  ? 

51.  A  boy  spent  \  of  his  money  and  |-  a  cent  more,  then  \  of 
the  remainder  and  ^  a  cent  more,  then  i  of  what  he  had  left  and 
^  a  cent  more,  when  he  found  that  he  had  2  cents  remaining. 
How  much  had  he  at  first  ? 

52.  Five  boys  bought  a  boat,  agreeing  to  share  the  expense 
equally.  But  one  of  them  having  left  $  1  of  his  share  unpaid, 
each  of  the  others  had  to  pay  ^^  ^lore  than  one  fifth  of  the  ex- 
pense.    What  was  the  cost  of  the  boat  ? 

53.  A  sum  of  money  was  divided  among  A,  B,  C,  and  D  so 
that  A  received  \  as  much  as  all  the  others,  B  received  \  as  much 
as  all  the  others,  C  received  \  as  much  as  all  the  others,  and  D 
received  $  2800  less  than  A.     What  sum  did  each  receive  ? 

54.  In  an  alloy  of  90  ounces  of  silver  and  copper  there  are  6 
ounces  of  silver.  How  much  copper  must  be  added  that  10 
ounces  of  the  new  alloy  may  contain  ^  of  an  ounce  of  silver  ? 

55.  If  80  pounds  of  sea  water  contain  4  pounds  of  salt,  how 
much  fresh  water  must  be  added  that  45  pounds  of  the  new  solu- 
tion may  contain  If  pounds  of  salt  ? 

56.  An  officer,  attempting  to  arrange  his  men  in  a  solid  square, 
found  that  with  a  certain  number  of  men  on  a  side  he  had  34  men 
over,  but  with  1  man  more  on  a  side  he  needed  35  men  to  com- 
plete the  square.     How  many  men  had  he  ? 

Suggestion.  —  With  x  men  on  a  side,  the  square  contained  x^  men ;  with 
X  -I-  1  men  on  a  side,  there  were  places  for  (a;  +  1)2  men.  Since  the  number 
of  men  was  the  same  under  both  arrangements,  x^  +  34  =  (x  +  1)^  —  35. 

57.  A  regiment  drawn  up  in  the  form  of  a  solid  square  lost  60 
men  in  battle.  Afterward,  when  the  men  were  arranged  in  a 
solid  square  with  1  man  less  on  a  side,  it  was  found  that  there 
was  1  man  over.  How  many  men  were  there  in  the  regiment  at 
first? 


180  .         SIMPLE   EQUATIONS 

58.  A  regiment  drawn  up  in  the  form  of  a  solid  square  was 
reenforced  by  240  men.  When  the  regiment  was  formed  again  in 
a  solid  square,  there  were  4  more  men  on  a  side.  How  many 
men  were  there  in  the  regiment  at  first  ? 

59.  A  man  was  hired  for  40  days  under  the  following  condi- 
tions :  for  every  day  he  worked  he  was  to  receive  $  3  besides  his 
board,  while  for  every  day  he  was  idle  he  was  to  receive  nothing, 
but  was  to  be  charged  $  1.20  for  his  board.  If  at  the  end  of  the 
period  he  received  $  57,  how  many  days  did  he  work  ? 

60.  A  man  invested  $800,  a  part  at  6%  and  the  rest  at  5%. 
If  the  total  annual  interest  was  $  45,  how  much  did  he  invest  at 
each  rate  ? 

Sdggkstion.  —  Let  x  =  the  number  of  dollars  invested  at  6  %. 
Then,  800  —  aj  =  the  number  of  dollars  invested  at  5  % ; 

61.  A  man  has  f  of  his  property  invested  at  4%,  \  at  3%,  and 
the  remainder  at  2%.  How  much  property  has  he,  if  his  annual 
income  is  $  860  ? 

62.  A  man  put  out  $4330  in  two  investments.  On  one  of 
them  he  gained  12%,  and  on  the  other  he  lost  5%.  If  his  net 
gain  was  $  251,  what  was  the  amount  of  each  investment  ? 

63.  There  were  distributed  among  20  men  and  25  women  $  160 
in  such  a  way  that  the  sum  of  what  a  man  and  a  woman  received 
was  $  7.  How  much  did  the  men  receive,  and  how  much  did  the 
women  receive  ? 

64.  At  what  time  between  5  and  6  o'clock  will  the  hands  of  a 
clock  be  together  ? 

Solution 
Let         X  =  the  number  of  minute  spaces  that  the  minute  hand  travels 

after  5  o'clock  before  they  come  together. 
Then,    —  =  the  number  of  minute  spaces  that  the  hour  hand  travels  in 

the  same  time. 
Since  they  are  25  minute  spaces  apart  at  6  o'clock, 

«-  — =  25;    . 

12  '  ^ 

.'.  X  =  27^,  the  number  of  minutes  after  5  o'clock. 

65.  At  what  time  between  1  and  2  o'clock  will  the  hands  of  a 
clock  be  together  ? 


SIMPLE   EQUATIONS  181 

66.  At  what  time  between  6  and  7  o'clock  will  the  hands  of  a 
clock  be  together  ? 

67.  At  what  time  between  10  and  11  o'clock  will  the  hands  of 
a  clock  point  in  opposite  directions  ? 

68.  At  what  times  between  4  and  5  o'clock  will  the  hands  of  a 
clock  be  15  minute  spaces  apart  ? 

69.  When  after  9  o'clock  and  before  10  o'clock  will  the  hands 
of  a  clock  be  at  right  angles  to  each  other  ? 

70.  A  man  rows  downstream  at  the  rate  of  6  miles  an  hour 
and  returns  at  the  rate  of  3  miles  an  hour.  How  far  downstream 
can  he  go  and  return  within  9  hours  ? 

71.  At  the  rate  of  3  miles  an  hour  uphill  and  4  miles  an  hour 
downhill  a  woman  can  walk  60  miles  in  17  hours.  How  much 
of  the  distance  is  uphill,  and  how  much  is  downhill  ? 

72.  A  hare  pursued  by  a  hound  takes  4  leaps  while  the  hound 
takes  3 ;  but  2  of  the  hound's  leaps  are  equal  to  3  of  the  hare's. 
If  the  hare  has  a  start  equal  to  GO  of  her  own  leaps,  how  many 
leaps  must  the  hound  take  to  catch  the  hare  ? 

Solution 
Let  3  a;  =  the  number  of  leaps  taken  by  the  hound. 

Then,  4  x  =  the  number  of  leaps  taken  by  the  hare. 

Suppose  a  =  the  number  of  feet  in  one  leap  of  the  hare. 

q   « 

Then,  —  =  the  number  of  feet  in  one  leap  of  the  hound, 

—  X  3  a;  =  _ax  ^^^  jmmber  of  feet  the  hound  runs, 
2  2 

and  a  X  4  a;  =  4  aa;,  the  rmmber  of  feet  the  hare  runs. 

Since  the  hare  has  a  start  equal  to  60  times  a  feet,  or  60  a  feet,  the  hare 
runs  60  a  feet  less  than  the  hound. 


Therefore, 

Aax  =  ^  **     60  a. 

2 

Dividing  by  a. 

4«  =  — -60. 

2 

Therefore, 

X  =  120, 

and 

3  X  =  360,  the  number  of  leaps  taken  by  the  hound. 

182  SIMPLE   EQUATIONS 

73.  A  fox  is  70  leaps  ahead  of  a  hound  and  takes  5  leaps  while 
the  hound  takes  8 ;  but  3  of  the  hound's  leaps  equal  7  of  the 
fox's.     How  many  leaps  must  the  hound  take  to  catch  the  fox  ? 

74.  A  rabbit  makes  5  leaps -while  a  dog  makes  4 ;  but  3  of  the 
dog's  leaps  are  equal  to  4  of  the  rabbit's.  If  the  rabbit  has  a 
start  of  20  leaps,  how  many  leaps  will  each  take  before  the  rabbit 
is  caught  ? 

75.  A  hound  is  39  of  his  leaps  behind  a  rabbit  that  takes  7 
leaps  while  the  hound  takes  8.  If  6  leaps  of  the  rabbit  are  equal 
to  5  leaps  of  the  hound,  how  many  leaps  must  the  hound  take  to 
catch  the  rabbit  ? 

76.  A  wheelman  and  a  pedestrian  leave  the  same  place  at  the 
same  time  to  go  to  a  point  54  miles  distant,  the  former  traveling 
3  times  as  fast  as  the  latter.  The  wheelman  makes  the  trip  and 
returning  meets  the  pedestrian  in  6f  hours  from  the  time  they 
started.     What  is  the  rate  of  each  ? 

77.  If  1  pound  of  lead  loses  -^  of  a  pound,  and  1  pound  o£ 
iron  loses  -^-^  of  a  pound  when  weighed  in  water,  how  many  pounds 
of  lead  and  of  iron  are  there  in  a  mass  of  lead  and  iron  that 
weighs  159  pounds  in  air  and  143  pounds  in  water  ? 

78.  If  97  ounces  of  gold  weighs  92  ounces  when  it  is  weighed 
in  water,  and  21  ounces  of  silver  weighs  19  ounces  when  it  is 
weighed  in  water,  how  many  ounces  of  gold  and  of  silver  are 
there  in  a  mass  of  gold  and  silver  that  weighs  320  ounces  in  air 
and  298  ounces  in  water  ? 

79.  A  merchant  increases  his  capital  annually  by  \  of  it,  and 
at  the  end  of  each  year  takes  out  $  800  for  expenses.  At  the  end 
of  three  years,  after  taking  out  his  expenses,  he  finds  that  his 
capital  is  $  6325.     What  was  his  original  capital  ? 

80.  A  merchant  added  annually  to  his  capital  \  of  it,  and  at 
the  end  of  each  year  took  out  f  1000  for  expenses.  If  at  the  end 
of  the  third  year,  after  taking  out  the  last  $  1000,  he  had  f  of 
his  original  capital,  what  was  his  original  capital  ? 

81.  A  cistern  can  be  filled  by  one  pipe  in  20  minutes,  by 
another  in  15  minutes,  and  it  can  be  emptied  by  a  third  in  10 


SIMPLE  EQUATIONS  183 

minutes.  If  the  three  pipes  are  running  at  the  same  time,  how 
long  will  it  take  to  fill  the  cistern  ? 

82.  A  man  walked  from  A  to  B  at  the  rate  of  2  miles  an  hour, 
and  rode  back  at  the  rate  of  3^  miles  an  hour,  being  gone  13 
hours.     How  far  is  it  from  A  to  B  ? 

83.  An  express  train  whose  rate  is  40  miles  an  hour  starts 
1  hour  and  4  minutes  after  a  freight  train  and  overtakes  it  in 
1  hour  and  36  minutes.  How  many  miles  does  the  freight  train 
run  per  hour  ? 

84.  The  distance  from  Albany  to  Syracuse  is  148  miles.  A 
canal  boat  leaves  Albany  for  Syracuse,  moving  at  the  rate  of 

3  miles  in  2  hours ;  at  the  same  time  another  leaves  Syracuse  for 
Albany,  moving  at  the  rate  of  5  miles  in  4  hours.  How  far  from 
Albany  do  they  meet  ? 

85.  A  steamer  goes  5  miles  downstream  in  the  same  time  that 
it  goes  3  miles  upstream ;  but  if  its  rate  each  way  is  diminished 

4  miles  an  hour,  its  rate  downstream  will  be  twice  its  rate  up- 
stream.    How  fast  does  it  go  in  each  direction  ? 

86.  A,  B,  and  C  together  can  do  a  piece  of  work  in  3^  days ; 
B  can  do  \  as  much  as  A,  and  C  can  do  |  as  much  as  B  in  a  day. 
In  how  many  days  can  each  do  the  work  alone  ? 

87.  A  can  do  a  piece  of  work  in  6  days  that  B  can  do  in  14 
days.  A  began  the  work,  and  after  a  certain  number  of  days  B 
took  his  place  and  finished  the  work  in  10  days  from  the  time 
it  was  begun  by  A.     How  many  days  did  B  work  ? 

88.  In  a  certain  weight  of  gunpowder  the  niter  composed  10 
pounds  more  than  |  of  the  weight,  the  sulphur  3  pounds  more 
than  y^2)  3,nd  the  charcoal  3  pounds  less  than  ^V  ^^  ^^®  niter. 
What  was  the  weight  of  the  gunpowder  ? 

89.  A  library  containing  16,000  volumes  was  divided  into  five 
departments.  In  the  department  of  history  there  were  twice  as 
many  volumes  as  in  the  department  of  science,  and  500  more 
than  \  as  many  volumes  as  in  the  juvenile  department.  Of 
fiction  there  were  1\  times  as  many  volumes  as  of  science,  and 
500  less  than  8  times  as  many  as  in  the  reference  department. 
How  many  volumes  were  there  in  each  department  ? 


184  SIMPLE   EQUATIONS 

90.  An  estate  was  divided  among  four  heirs,  A,  B,  C,  and  D. 
If  the  value  of  the  estate  had  been  $  1000  less,  what  A  received 
would  have  been  \  of  it,  and  what  B  received  ^  of  it ;  if  the  value 
of  the  estate  had  been  \  greater,  what  C  received  would  have 
been  \  of  it,  and  what  D  received  \  of  it.  What  sum  did  each 
receive  ? 

91.  A  father  takes  3  steps  while  his  son  takes  5 ;  but  2  of  the 
father's  steps  are  equal  to  3  of  the  son's.  How  many  steps  will 
the  son  require  to  overtake  his  father,  who  is  36  steps  ahead  ? 

92.  A  purse  contained  some  money  and  a  ring  worth  $10 
more  than  the  money.  If  the  purse  was  worth  ^  as  much  as  the 
money  it  contained,  g,nd  the  purse  and  the  money  together  were 
worth  \  as  much  as  the  ring,  what  was  the  value  of  each  ? 

93.  Brass  is  8|  times  as  heavy  as  water,  and  iron  1\  times  as 
heavy  as  water.  A  mixed  mass  weighs  57  pounds,  and  when 
immersed  displaces  7  pounds  of  water.  How  many  pounds  of 
each  metal  does  the  mass  contain  ? 

94.  A  man  began  business  with  $4725,  and  annually  added  to 
his  capital  4  of  it.  At  the  end  of  each  year  he  put  aside  a 
certain  sum  for  expenses.  If  at  the  end  of  the  third  year,  after 
taking  out  the  sura  for  expenses,  his  capital  was  $3800,  what 
were  his  annual  expenses  ? 

95.  The  sum  of  two  numbers  is  s,  and  their  difference  d. 
What  are  the  numbers  ? 

Solution 
Let  X  =  the  greater  number. 

Then,  X  —  (Z  =  the  less  number, 

and  X  +  X  -  d  =  s  ; 

.-.  X  =  ^"*"    ,  the  greater  number,  (1) 

and  —  d  =  - — -,  the  less  number.  (2) 

If  the  sum  of  two  numbers  is  30,  and  their  difference  is  6, 
what  are  the  numbers  ? 

By  (1),  the  greater  number  is  — ^,  or  18  ; 

by  (2),  the  less  number  is  '- — ^^—i  or  12. 


SIMPLE   EQUATIONS  185 

A  problem  in  which  the  numbers  assumed  to  be  known  are 
represented  by  letters  to  which  any  values  may  be  assigned  is 
called  a  General  Problem. 

Problem  95  is  a  general  problem. 

The  results  obtained  in  solving  a  general  problem  may  be  con- 
sidered formvlce  for  solving  similar  problems. 

96.  Divide  c  cents  between  two  boys  so  that  one  shall  have  d 
cents  more  than  the  other.  If  c  —  50  and  d  =  10,  how  much  will 
each  receive  ? 

97.  A  horse  and  a  saddle  are  together  worth  a  dollars,  and  the 
horse  is  worth  m  times  as  much  as  the  saddle.  What  is  the  value 
of  each  ?     What,  if  a  =  160  and  m  =  3  ? 

98.  Divide  6  into  two  parts  one  of  which  represents  m  times 
the  other.     What  will  they  be,  if  b  represents  100,  and  m,  4  ? 

99.  An  estate  of  a  dollars  is  divided  between  two  heirs  in  the 
proportion  of  m  to  n.  What  is  the  share  of  each  ?  What  is  the 
share  of  each,  if  a  =  40,000,  m  =  5,  and  ?t  =  3  ? 

100.  If  A  can  do  a  piece  of  work  in  a  days,  and  B  in  6  days, 
in  what  time  can  both  do  it  working  together  ?  Give  the  result, 
if  a  =  10  and  b  =  15. 

101.  An  alloy  of  two  metals  is  composed  of  m  parts  of  one  to 
n  parts  of  the  other.  How  many  pounds  of  each  are  required  in 
the  composition  of  a  pounds  of  the  alloy  ? 

Bell  metal  is  an  alloy  of  5  parts  of  tin  and  16  parts  of  copper. 
How  many  pounds  of  tin  and  of  copper  are  there  in  a  bell  which 
weighs  4200  pounds  ? 

102.  A  wheelman  set  out  from  B  at  the  rate  of  r  miles  an  hour. 
a  hours  later  another  started  in  pursuit  at  the  rate  of  p  miles  an 
hour.  How  far  from  B  will  the  second  wheelman  overtake  the 
first  ?     What  will  be  the  distance,  if  r  =  10,  p  =  12,  and  a  =  8  ? 

103.  A  man  traveled  from  home  at  the  rate  of  a  miles  an  hour 
and  returned  at  the  rate  of  b  miles  an  hour.  If  he  made  the 
entire  journey  in  h  hours,  how  far  from  home  did  he  go  ?  How 
far,  if  a  =  4,  &  =  3^,  and  h  =  15? 


SIMULTANEOUS    SIMPLE   EQUATIONS 


TWO  UNKNOWN  NUMBERS 

201.  1.  It  x  +  y=12,  what  is  the  value  of  x?  oty?  How 
many  values  may  x  have  ?     How  many  may  y  have  ? 

2.  In  the  expression  x  -\-  y  =  12,  x  and  y  each  may  have  an 
indefinite  number  of  values,  but  if,  at  the  same  time,  x  —  y  =  4, 
what  is  the  value  of  a;  ?  of  y? 

3.  Although  one  equation  containing  two  unknown  numbers 
has  an  indefinite  number  of  values  for  each  unknown  number,  or 
is  indeterminate,  what  can  be  said  about  the  values  of  the  un- 
known numbers,  when  two  equations  are  given  involving  the  same 
values  of  the  unknoAvn  numbers,  but  in  difEerent  relations,  that  is, 
when  two  independent  equations  are  given  ? 

202.  Two  or  more  equations  in  which  the  unknown  numbers 
have  the  same  values  are  called  Simultaneous  Equations. 

If  X  and  y  represent  the  same  numbers  in2a;-t-3?/  =  19as  they  represent 
inSx  —  j/=22,  2a:  +  32/  =  19  and  5  a;  —  y  =  22  are  simultaneous  equations. 

203.  Equations  that  represent  different  relations  between  the 
unknown  numbers,  and  so  cannot  be  reduced  to  the  same  form,  are 
called  Independent  Equations. 

3x  +  Sy  =  IS  and  2 x  +  2 1/  =  12  really  express  but  one  relation  between 
X  and  y  ;  viz.,  that  their  sum  is  6.  Hence,  both  equations  may  be  reduced  to 
the  same  form,  as  x  +  y  =  6.  But  3  x  +  3  v  =  18  and  x  +  Sy  =  14  express 
different  relations  between  x  and  y,  and  cannot  b«  reduced  to  the  same  form. 
Hence,  they  are  independent  equations. 

204.  An  equation  whose  unknown  numbers  may  have  an  in- 
finite number  of  values  is  called  an  Indeterminate  Equation. 

X  +  1/  =  6  is  an  indeterminate  equation,  because,  if  x  =  2,  y  —  4 ;  if  «  =  3, 
y  =  3 ;  if  X  =  2|,  J/  ^  3^ ;  etc. 

186 


SIMULTANEOUS  SIMPLE  EQUATIONS  187 

205.  Principle.  —  Every  single  equation  involving  two  or  more 
unknown  numbers  is  indeterminate. 

206.  The  process  of  deriving  from  a  set  of  simultaneous  equa- 
tions, equations  involving  a  less  number  of  unknown  numbers 
than  is  found  in  the  given  equations  is  called  Elimination. 

207.  Two  sets  of  simultaneous  equations  each  having  all  the 
roots  of  the  other  set  are  called  Equivalent  Systems  of  equations. 

Thus  I       x  +  y  =  %    \      ^     r2x  +  2y  =  16| 

Itius,  \  3  a; +  2  2/ =  21/    ^^"^    \3x  +  2?/  =  2l| 

are  equivalent  systems,  for  each  system  is  satisfied  by  the  same  set  of  values, 
X  =  5  and  ?/  =  3,  and  neither  is  satisfied  by  any  other  set  of  values. 

208.  Any  equation  in  a  set  of  simultaneous  equations  may  be 
transformed  into  an  equivalent  equation  by  employing  the  prin- 
ciples of  equivalency  stated  in  §  196,  since  these  principles  apply 
to  all  equations.  Hence,  it  remains  to  seek  the  principle  by  which 
simultaneous  equations  are  combined  in  the  process  of  elimination 
without  introducing  or  losing  roots. 

This  Principle  of  Elimination  may  be  stated  as  follows,  a  and  h 
being  known  multipliers,  not  zero,  and  either  positive  or  negative : 

If  any  equation'  of  a  system  is  replaced  by  the  stim  or  difference 
of  a  times  that  equation  and  b  times  another  equation  of  the  system, 
the  resulting  system  is  equivalent  to  the  given  system. 

The  proof  is  as  follows : 

By  §  196,  Prin.  1,  all  the  terms  of  the  second  member  of  an  equation  may 
be  transposed  to  the  first  member. 


Then,  let  ^  =  0 

B  =  0 


(1) 


be  the  given  system,  and  let  a  and  6  be  any  knovfn  multipliers  except  zero 
and  either  positive  or  negative. 
It  is  to  be  proved  that  the  system 

aA  +  bB  =  0^ 

5  =  0  (2) 


is  equivalent  to  the  given  system  (1). 

Since  a  and  b  are  knovsrn  multipliers,  not  zero,  by  §  196,  Prin.  2,  every  set 
of  values  of  the  unknown  numbers  that  makes  ^  =  0  makes  aA  =  0,  and 
every  set  of  values  that  makes  .6  =  0  makes  bB  =  0.     Hence,  every  set  of 


188  SIMULTANEOUS   SIMPLE  EQUATIONS 

values  that  satisfies  (1)  makes  rt^  +  hB  equal  to  zero  and  thus  satisfies  (2), 
since  all  of  the  equations  of  (1)  and  (2)  except  the  first  are  the  same. 

Again,  since  every  set  of  values  that  satisfies  (2)  makes  B,  or  bB  equal  to 
zero  and  also  makes  aA  +  hB  =  0,  every  such  set  of  values  makes  aA  =  0 
and  therefore,  by  §  196,  Prin.  2,  makes  ^  =  0  and  satisfies  (1). 

Since  all  the  roots  of  (1)  are  roots  of  (2)  and  all  the  roots  of  (2)  are  roots 
of  (1),  (1)  and  (2)  are  equivalent  systems. 

209.    Elimination  by  Addition  and  Subtraction. 

1.  It  2 X  +  2y  =  10  and  3x  —  2 y  =  5  are  added,  what  is  the 
resulting  equation  ?  When  may  an  unknown  number  be  elimi- 
nated by  addition  ? 

2.  If  a;  +  2  ?/  =  5  is  subtracted  from  3  x  -\-  2  y  =  11,  what  is 
the  resulting  equation  ?  When  may  a  number  be  eliminated  by 
subtraction  ? 

Principle.  —  A  literal  number  having  the  same  coefficient  in 
two  equations  may  be  eliminated  by  adding  the  equations,  if  the 
coefficients  have  unlike  signs,  or  by  subtracting  one  equation  from  the 
other,  if  the  coefficients  have  like  signs. 

Examples 

1.   Find  the  value  of  x  and  of  ?/  in  2  x+3  y  =  7  and  3  x+4:  y=10. 

Explanation.  —  Since  x  has  not  the  same  co- 
eflBcients   in    both   equations,   the  equations  are 

2  X  4-  3y  =    7       (1)      multiplied  by  such  numbers  as  will  make  the  co- 

3  a;  -4-  4  V  =  10       ^2^      efficients  of  x  alike.     Multiplying  eq.  (1)  by  3  and 

eq.  (2)  by  2  gives  equations  (3)  and  (4).     From 

6  X  -{-  2  y  =  21  (3)  these  x  is  eliminated  by  subtraction,  and  the  value 
6  a;  +  8  ?/  =  20       (4)      of  y  is  found. 

The  value  of  x  may  be  found  by  multiplying 

2/  ~    ■*■       \^)      eq.   (1)  and  (2)  by  such  numbers  as  will  make 
2  a;  +  3  =    7       (6)      the  coefficients  of  y  equal  and  then  subtracting 
X  =    2       (7)      the  resulting  equations. 

Or,  the  value  of  x  may  be  found  by  substituting 
the  value  of  y  for  y  in  one  of  the  given  equations,  as  eq.  (1). 

Proof  of  the  Equivalence.  —  By  the  principle  of  elimination-,  the  system 
(1, 6),  obtained  by  subtracting  (4)  from  (3),  is  equivalent  to  the  given  system 
(1,  2).     Hence,  the  only  value  of  y  in  the  given  system  is  1. 

Since  y  represents  1  in  the  given  system,  (1)  is  equivalent  to  (6),  which 
by  §  196  is  equivalent  to  (7).  Hence,  the  given  system,  being  equivalent  to 
(1,  5),  is  equivalent  to  (7,  5),  whicj^  is  satisfied  by  one  and  only  one  set  of 
values,  X  =  2  and  y  =  1. 


SIMULTANEOUS  SIMPLE   EQUATIONS 


189 


Rule.  —  If  necessary,  multiply  the  equations  by  such  numbers  as 
will  cause  the  coefficients  of  one  letter  to  be  numerically  equal  in  the 
resulting  equations. 

When  the  signs  of  these  coefficients  are  unlike,  add  the  equations; 
when  the  signs  are  alike,  subtract  one  equation  from  the  other. 


10. 


11. 


12. 


Solve  by  addition  or  subtraction ; 

7  a;  —  5  y  =  52, 

2  a;  +  5  y  =  47. 

3  a?  +  2  2/  =  23, 

x  +  y  =  ^. 

3  .r  -  4  ?/  =  7, 

x  +  10?/  =  25. 

r  2  a;  -  10  2/  =  15, 
2a;-4y  =  18. 

(-  3  a;  -  y  =  4, 
U+3i/  =  -2. 

4  a;  —  y  =  19, 
a;  +  3  y  =  21. 

a;  +  2  2/  =  5, 
2x  +  y  =  l. 

2x  +  3y  =  17, 

3  a;  +  2  ?/  =  18. 

3  a:  -f  4  7/  =  25, 
4.x-\-Sy  =  31. 

5x  +  6y  =  32, 
7x-Sy  =  22. 

3x  +  6y  =  S9, 
9  a;  —  4  2/  =  51. 


13, 


14. 


15. 


16. 


17. 


18. 


19. 


20. 


(7x-9y  =  6, 

[x  +  2y  =  U. 

13x-y  =  20, 
4:x  +  2y  =  20. 

(Sx-3y^U, 
\7x-5y  =  29. 

6x-5y  =  33, 
4  a;  +  4  y  =  44. 

a;  +  14  y  =  38, 
14  a;  +  y  =  142 

5x  +  y  =  12, 
X  -\-  5  y  =  36. 

3  a; +  11?/ =  67, 
5x  —  3y  =  b. 

-  +  ^  =  12, 

4  2         ' 


|  =  _2. 


21.    < 


3     3 


7, 


+  ^  =  6i 


X 

6^2 


190  SIMULTANEOUS   SIMPLE   EQUATIONS 

210.   Elimination  by  Comparison. 

1.  If,  in  the  simultaneous  equations  x  —  y  =  ^  and  a;  +  4  ?/  =13, 
the  terms  containing  y  are  transposed,  what  will  be  the  resulting 
equations  ? 

2.  Since  the  second  members  of  these  derived  equations  are 
each  equal  to  x,  how  do  they  compare  with  each  other  ? 

3.  If  these  second  members,  then,  are  placed  equal  to  each 
other,  how  many  itnknown  numbers  Avill  this  equation  contain  ? 

4.  How  may  an  unknown  number  be  eliminated  from  two 
simple  simultaneous  equations  by  comparison? 

Examples 
1.   Find  the  value  of  x  and  of  2/  in  2  x—Z  ;y=10  and  5  a;+2  ?/=6. 

Explanation.  —  Solving  eq.  (1)  as  if 
X  were  the  only  unknown  number,  the 
value  of  X  is  given  as  in  eq.  (3).  In  like 
manner,  solving  eq.  (2)  for  x,  another 
expression  is  found  for  the  value  of  x,  as 
in  eq.  (4). 

Since  the  equations  are  simultaneous, 

6  —  2y        ^ .  X       the    unknown    numbers    have  the  same 

/•^   yy?  g  ^  /       values  in  each,  and  the  two  values  of  x 

^(\       "X  k       O  form  an  equation,  as  eq.  (5),  from  which 

.'.  — — ^  = ^        (5)      X  has  been  eliminated.     Solving  eq.  (5), 

^  '^  the  value  of  y  is  found  to  be  —  2,  eq.  (6). 

.*.  2/  =  —  2  (6i)  By  a  similar  process  the  value  of  X  may 

10  —  6  /Q\       ^6  found  ;  or,  substituting  the  value  of  y 

2  foj"  y  ^^  0116  of  the  preceding  equations, 

_  2  /j\      as  (.3),  the  value  of  x  is  found  to  be  2, 

~  eq.  (7). 

Proof  of  the  Equivalence.  — By  §  196,  (3)  is  equivalent  to  (1)  and  (4) 
is  equivalent  to  (2).  Hence,  the  system  (3,  4)  is  equivalent  to  the  given 
system. 

Since  (5),  which  by  §  196  is  equivalent  to  (6),  may  be  obtained  by  sub- 
tracting (3)  from  (4)  and  transposing,  by  the  principle  of  elimination,  §  208, 
(6,  1).  or  (6,  3),  is  equivalent  to  (4,  3),  or  to  (2,  1),  the  given  system. 

But  since  —  2  is  the  only  value  of  y  in  the  given  system,  by  §  196,  (7)  is 
equivalent  to  (3),  or  to  (1).  Hence.  (7,  6),  which  is  satisfied  by  one  and 
only  one  set  of  values,  x  =  2  and  y  =  —  2,  is  equivalent  to  the  given  system. 
Therefore,  the  given  system  is  satisfied  by  x  =  2,  y  =  —  2  and  by  no  other 
values  of  x  and  y. 


PROCESS 

2x 

-3y  = 

:10 

(1) 

5x 

+  2y  = 

:      6 

(2) 

X  = 

10+32/ 
2 

(3) 

SIMULTANEOUS  SIMPLE  EQUATIONS 


191 


KuLE.  —  Find  an  expression  for  the  value  of  the  same  unknown 
number  in  each  equation,  equate  the  two  expressions,  and  solve  the 
equation  thus  formed. 


Solve  by  comparison : 
[x  +  y  =  70. 


(5x  +  y  =  22, 


[X  +  5y  =14:. 


4.     \ 


(2x  +  3y  =  24:, 


[5x-3y=lS. 

(3x  +  5y=14:, 
'     {2x-3y=3. 

g      (3x  +  2y==36, 
\5x-9y  =  23. 


7.     \ 


l2x  +  7y  =  S, 


[3x  +  9y  =  9. 


g      Ux  +  6y  =  19, 
\3x-2y=^^. 


10.     \ 


4a; +  3?/ =  34, 

11  a; +  5?/ =  87. 

(4:x-13y  =  5, 


11. 


13. 


14. 


15. 


[3x-\-lly=-17. 

lSx-3y  =  Ay, 
1- 4a; +  3?/ =  27. 


12.     i 


(7y-X  =  0, 


\x  +  2y  =  lS. 

(3y  +  9  =  5x, 
|l6-2a;  =  5y. 

5  a;  —  40  =  y, 

5y  —  60  =  X. 

(2y-llx  =  67, 

I  2a; +  5?/ =20. 


211.   Elimination  by  Substitution. 

1.  In  an  equation  containing  two  unknown  numbers,  if  the 
value  of  one  is  found,  how  is  the  value  of  the  other  obtained  ? 

2.  Express  the  value  of  x  in  the  first  of  the  simultaneous 
equations  x  +  y  =  5  and  x  +  2y  =  7  by  transposing  y  to  the 
second  member. 

3.  When  this  expression  for  the  value  of  x  is  substituted  for 
X  in  the  second  equation,  how  many  unknown  numbers  does  the 
resulting  equation  contain  ? 

4.  How  may  an  unknown  number  be  eliminated  from  simul- 
taneous equations  by  substitution  ? 


PKOCESS 

X 

+  5y. 

=  9 

(1) 

3a; 

-2y. 

=  10 

(2) 

X 

=  9- 

6y 

(3) 

3(9- 

-52/) 

-2y  = 

=  10 

(4) 

.'.  2/  = 

=  1 

(5) 

x  = 

=  9- 

■5 

(3) 

192  SIMULTANEOUS   SIMPLE   EQUATIONS 

Examples 
1.    Find  the  value  of  x  and  of  y  in  ic+r>?/=9  and  3x—2y=10. 

Explanation.  —  Solving  eq.  (1) 
for  x,  X  =  9  —  5  y. 

Since  the  given  equations  are 
simultaneous,  x  has  the  same  value 
in  eq.  (2)  as  in  eq.  (1). 

If  9  —  5  ?/  is  substituted  for  x  in 
eq.  (2),  the  resulting  equation  will 
be  true.     Substituting  and  solving, 

y  =  l. 

Substituting  the  value  of  y  for  y 
x  =  4  (6)  in  eq.  (3),  x  =  4,  eq.  (6). 

Proof  of  the  Equivalence.  — By  §  196,  (3)  is  equivalent  to  (1).  In  the 
process,  (4)  is  obtained  by  substituting  the  expression  equal  to  x  in  (3)  for  x 
in  (2),  or  by  substituting  3(9  —  5  ?/)  for  3x  in  (2).  Since  the  substitution 
of  3(9  —  5  y)  for  3  x  may  be  performed  by  subtracting  3  a;  =  3(9  —  5 «/)  from 
(2),  by  the  principle  of  elimination  (4),  or  the  equivalent  equation  (5),  may 
take  the  place  of  (2)  in  the  system  (3,  2)  equivalent  to  the  given  system. 
Hence,  (3,  6)  is  equivalent  to  (1,  2). 

Since  the  only  value  of  y  in  the  system  (3,  5)  is  1,  (6)  is  equivalent  to  (3); 
and  (6,  5),  which  is  satisfied  by  only  one  set  of  values,  a;  =  4,  y  =  1,  is  equiva- 
lent to  (3,  5)  and  therefore  to  (1,  2). 

Rule.  —  Find  an  expression  for  the  value  of  one  of  the  unknown 
mimbers  in  one  of  the  equations. 

Substitute  this  value  for  that  unknown  number  in  the  other  equof 
tion,  and  solve  the  resulting  equation. 

Solve  by  substitution : 

2      \x-y  =  4:,  ri7  =  3a;-f2, 

\4:y-x^U.  '     \l  =  3z-2x. 

^      |x  +  2/  =  10,  g_     {4.y  =  \0-x, 

[6a;— 72/  =  34.  {y  —  x  =  b. 


4. 


6. 


3a;-4?/  =  26,  ^      r7z-3x  =  18, 
x-^y  =  22.  '     \2z-bx  =  l. 

62/-10a;  =  14,  ^^      r3-152/=-a;, 
y  —  x  =  3.  {  3  + 15  ?/  =  4  a;. 

(y  +  l  =  3x,  ^^      (l-x  =  3y, 

[5x  +  9  =  3y.  '     1  8(1- a;)  =  40-2/. 


SIMULTANEOUS   SIMPLE  EQUATIONS 


193 


Solve  by  any  method,  eliminating  without  clearing  of  fractions, 
when  possible : 


12. 


x  +  |  =  ll, 


+  3y  =  21. 


19. 


3  4 


13. 


^  +  ^  =  21, 
4        5 

3        5 


20. 


2     3  ' 

2a;-l    .3y-1^5 
2  3  6* 


fa; 


14.   •! 


3  2 

3^  7 


21. 


'  1  +  x      2a;  — «      o         E? 
5y^  +  i^3^18-5ar. 


15. 


^f  +  42/  =  15, 


6       3 


22. 


5  _  12  =  ^4-  8, 
2  4       ' 

^  +  y  ,  a; _ 2y-a;      ^k 


16. 


fa;-! 
4 

a;-l 

I     4 


+  2/  =  3, 
+  4y  =  9. 


23. 


f      1 


x  —  1      a;  +  y 
3 


=  0, 


x-y 


+  3  =  0. 


17. 


3  a;      2  .V  _  9^ 
^  +  -3--20, 


24. 


^     1 9  ^  y  +  32 
2        "  4     ' 

y      3a;-2y_o- 
8"^^5~~-'^^- 


18. 


^  _y 

3~2' 

?_^  =  1 
3     3 

ALG.  —  13 


25. 


■2  y  +  .5  ^  .49  a;  -  .7 

1.5  4.2      ' 

.5  a; -.2  ^41      1.5y-ll 

1.6  16  8        ' 


fA 


194 


SIMULTANEOUS  SIMPLE  EQUATIONS 


26. 


27. 


28. 


29. 


30. 


31. 


a;  +  i(3a;  -  2/  -  1)  =  ^  +  |(y  _  1), 
|(4a;  +  32/)  =  J^(72/  +  24). 

'  6^  +  9  ,  3^  +  5  V      oi    ,  3a;-h4 

Ly  +  7      6a; -3y^        4.^-9 
10      "^2(y-4)         "^      5 

3^-5.v     2a;-8y-9     31 
3  12  ~  12' 


f+!+^^ 


0-( 


4  a;  -  2^  -  25 


^-.9n_2y-a;_2a;-59 
23  -  a;  2 

y-lziy_3o=gy-73. 

a;  -  18  3 

3a;  +  6       x-\-5       Qx  —  2 


3y-5 


14 


32.     ^ 


2y-3      3y  +  4^3y  +  5 
6       "^5a;-7  9 

'a;     16  —  a;      on  ,  5w  +  2a; 

2y-3  ,  83-8y     .^ 
yf8-  +  "8       =^'-^- 


23/- 


4a;4- 


33.     i 


17 -3  a; 


2       ^      16a;-l 


50        y-^    -9.,  I  147-24y 


SIMULTANEOUS  SIMPLE   EQUATIONS 


195 


34.    Solve 


U 

X 


14 


2     5^25^ 
!.«     y      3 


SOLCTIOK 

4_  3  _14 

X      y       b 

2       5  ^25 
X       2/       3 


(2)  X  2, 
(3)-(l), 


4      10, 
«       J/ 

13 


60 
3 

208 
16 

16 
16 


^      16 
Substituting  the  value  of  -  in  equation  (1), 


48. 
15" 

1^ 

X 


14 
6 

3 
2 


x=?. 
3 


(1) 

(2) 
(3) 
(4) 
(5) 
(6) 

(7) 
(8) 
(9) 


In  fractional  equations  in  which  the  denominators  are  simple  expressions 
like  the  above,  an  unknown  number  should  be  eliminated  before  clearing  of 
fractions. 


35. 


36. 


2  3  ' 

^4-^  =  12. 
2       3 

^  +  ^  =  13, 

3  3         ' 

f  "3  =  -^- 


37. 


38. 


-  +  -  =  64, 

X     y 

^  +  5  =  73^. 
X      y 

X      y 
X      y 


196 


SIMULTANEOUS  SIMPLE   EQUATIONS 


39. 


40. 


41. 


r5    3 


X 


y 


=  -% 


2^  +  1  =  6. 
X      y 


2     3. 

=  5, 

X     y 


X     y 

4,3      9 

X     y      o 

3      4^11 
X     y      12' 


42.     -! 


43. 


44. 


i  +  5  =  30, 

X      y 

1  +  ^  =  30. 

2x     y 

—  +  -  =  23. 
2a;     y 

8j;     3y         ' 


0 

6cc 


lly 


=  17. 


LITERA.L  Simultaneous  Equations 


1.    Solve  \<^  +  'by  =  ^, 
Kcx  +  dy  =  n. 


(1)  X  d, 

(2)  X  b, 
(3) -(4), 


(l)xc, 
(2)  X  a, 
(7) -(6), 


Solution 
ax  +  6y  =  m 
ex  +  dy  =  n 
adx  +  6^2/  =  dm 

hex  +  bdy  =  bn 
{ad  —  6c)a:  =  dm  —  bn 

.       _  dm  —  &» 
ad  —  be 


acx  +  fecy 
acx  +  ady 


cm 
an 


(ad  —  bc)y  ■■ 

■'■  y 


an 


cm 


ad  —  be 


(1> 
(2) 
(3) 

(4) 

(5) 

(6) 
(7) 

(8) 


In  literal  simultaneous  equations,  elimination  is  usually  performed  by  the 
method  of  addition  and  subtraction. 


SIMULTANEOUS  SIMPLE   EQUATIONS 


197 


3. 


5. 


6. 


9. 


10. 


11. 


■  ax  +hy  =  m, 
bx  —  ay=:c. 

f  ax  —  by  =  m, 
[ex  —  dy  =  r. 

f  ax  =  by, 
\x  +  y  =  ab. 

( X  —ay  =  n, 
[  bx  +  y  =  p. 

(a(x-y)  =  5, 
[  bx  —  cy  =  n. 

(  a(a  —  x)  =  b(y^b), 
\ax=  by. 

\x  +  y  =  b—a, 
\bx  —  ay  +  2ab  =  0. 

x     y     a 
11^1 

( X     y      b 


X     y 

b  _«__i 

X     y 


a     b 

l  ab     ab 


12. 


i  a      b 

[  bx  —  ay  =  0. 


13. 


14.    •( 


a     b 
6     a~2 


1    .   1 

— +  r-  =  c> 
ax     by 


15. 


6a;     ay 


a;      y 


ra;4-l_a  +  6  +  l 
16.    •(2/  +  1      a-b  +  1 
[x-y  =  2b. 


x-{-y_x  —  y 


17. 


18. 


a 

& 

x 

-y  - 

a 
1 

L 

1_ 

a 

X 

—  a 

a 

— 

y 

X 

+  2/_ 

a 

19.    \ 


20. 


a;-?/ 


a     b 
be 


+ 


_a 


a  +  a;      6  —  2/     & 
6       ,      a 


.a  +  X     b  —  y     a 


198  SIMULTANEOUS  SIMPLE  EQUATIONS 

Problems 

1.  There  are  two  numbers  such  that  if  twice  the  first  is  added 
to  3  times  the  second,  the  sum  will  be  130 ;  but  if  5  times  the 
first  is  diminished  by  the  second,  the  remainder  will  be  70. 
What  are  the  numbers  ? 


Solution 

Let 

X  —  the  first  number, 

and 

y  =  the  second  number. 

Then, 

2  a;  +  3  y  =  130, 

and 

5  X  -    2/  =  70. 

Eliminating  y, 

17  x  =  340, 

a;  =  20. 

Whence,  by  substitution 

y  =  30. 

2.  A  drover  sold  3  cows  and  7  horses  to  one  person  for  $  600, 
and  to  another  person,  at  the  same  prices,  3  cows  and  3  horses  for 
$  300.     How  much  per  head  did  he  get  for  each  ? 

3.  With  $  30  a  man  can  buy  20  yards  of  one  kind  of  cloth  and 
50  yards  of  another ;  with  $  23  he  can  buy  30  yards  of  the  first 
kind  and  20  yards  of  the  second  kind.  What  is  the  price  of  each 
per  yard  ? 

4.  If  45  bushels  of  wheat  and  37  bushels  of  rye  together  cost 
$  62.70,  and  37  bushels  of  wheat  and  25  bushels  of  rye,  at  the 
same  prices,  cost  $  48.30,  what  is  the  price  of  each  per  bushel  ? 

5.  Henry  expended  95  cents  for  apples  and  oranges,  paying 
5  cents  for  each  orange  and  4  cents  for  each  apple.  If  he  had  22 
of  both,  how  many  of  each  did  he  buy  ? 

6.  Five  years  ago  A  was  ^  as  old  as  B,  and  10  years  hence  he 
will  be  ^  as  old  as  B..    What  are  their  ages  ? 

7.  A  said  to  B,  "  If  you  were  twice  as  old,  and  I  were  ^  as  old, 
or  if  you  were  ^  as  old,  and  I  were  3  times  as  old,  the  sum  of 
our  ages  would  be  70."     How  old  was  each  ? 

8.  A  boy  is  given  28  cents  to  buy  a  dozen  cakes.  He  finds 
that  some  cost  2  cents  each  and  some  3  cents  each.  How  many 
of  each  kind  can  he  purchase  ? 


SIMULTANEOUS   SIMPLE   EQUATIONS  199 

9.  A  said  to  B,  "  Give  me  $  20,  and  I  shall  have  3  times  as 
much  money  as  you."  B  replied,  "  Give  me  $  5,  and  I  shall  have 
twice  as  much  money  as  you."     How  much  money  had  each  ? 

Solution 

Let  -x  =  the  number  of  dollars  A  had, 

and  y  =  the  number  of  dollars  B  had. 

Then,  a;  +  20  =  3(y  -  20), 

and  y+    5  =  2(x—    5). 

Solving,  X  =  25,  the  number  of  dollars  A  had, 

and  y  =  35,  the  number  of  dollars  B  had. 

10.  If  A  gives  B  $  100,  B  will  have  4  times  as  much  money  as 
A ;  but  if  B  gives  A  $  200,  A  will  have  4  times  as  much  money 
as  B.     What  sum  of  money  has  each  ? 

11.  A  said  to  B,  "  Give  me  20  cents  of  your  money,  and  I  shall 
have,  half  as  much  as  you."  B  replied,  "Give  me  25  cents  of 
your  money,  and  I  shall  have  5  times  as  much  as  you."  How 
much  had  each  ? 

12.  If  A  had  $300  more,  he  would  have  twice  as  much  as  B; 
if  B  had  $  300  less,  he  would  have  \  as  much  as  A.  How  much 
money  has  each  ?  • 

13.  If  1  is  added  to  each  term  of  a  fraction,  its  value  will 
be  f ;  if  1  is  subtracted  from  each  term  of  the  fraction,  its  value 
will  be  ^.     What  is  the  fraction  ? 

Solution 
Let  -  represent  the  fraction. 

y 

Then,  =«=+l-2 


and 


y+1     3' 

x-l     1 


y-1     2 
Solving,  a;  =  3, 

and  y  =  5, 

3 

That  is,  -  is  the  fraction. 

6 


200  SIMULTANEOUS  SIMPLE   EQUATIONS 

14.  If  1  is  added  to  the  numerator  of  a  certain  fraction,  its 
value  becomes  f ;  if  2  is  added  to  the  denominator,  its  value 
becomes  ^.     What  is  the  fraction  ? 

15.  Find  a  fraction  that  is  equal  to  \  when  its  terms  are 
diminished  by  2,  and  is  equal  to  f  when  its  terms  are  increased 
by  2. 

16.  A  certain  number  expressed  by  two  digits  is  equal  to  7 
times  the  sum  of  its  digits ;  if  27  is  subtracted  from  the  number, 
the  difference  will  be  expressed  by  reversing  the  order  of  the 
digits.     What  is  the  number? 


SOLCTION 

Let 

05  =  the  digit  in  tens'  place. 

and 

y  =  the  digit  in  units'  place. 

Then, 

10  a;  +  y  =  the  number, 

and 

10  y  +  a;  =  the  number  with  its  digits  reversed ; 

.-.  10«  +  j^  =  7(x  +  y), 

and 

10  X  +  y  -  27  =  10 1/ +  X. 

Solving, 

x  =  6, 

and 

2/ =  3. 

Hence, 

10  a;  +  !/  =  60  +  3,  or  63,  the  number. 

17.  The  sum  of  the  two  digits  of  a  certain  number  is  12,  and 
the  number  is  3  greater  than  6  times  the  sum  of  its  digits. 
What  is  the  number  ? 

18.  When  a  certain  number  expressed  by  two  digits  is  divided 
by  the  sum  of  its  digits,  the  quotient  is  8 ;  and  when  the  first 
digit  is  diminished  by  3  times  the  second,  the  remainder  is  1. 
What  is  the  number? 

19.  The  sum  of  the  two  digits  of  a  number  is  12.  If  the 
order  of  the  digits  is  reversed,  the  number  will  lack  12  of  being 
doubled.     What  is  the  number  ? 

20.  A  farmer  bought  100  acres  of  land  for  $  3250.  If  part  of 
it  cost  him  $  40  an  acre  and  the  rest  of  it  ^  15  an  acre,  how  many 
acres  were  there  of  each  kind  ? 


SIMULTANEOUS  SIMPLE  EQUATIONS  201 

21.  The  admission  to  an  entertainment  was  50  cents  for  adults 
and  35  cents  for  children.  If  the  proceeds  from  100  tickets 
amounted  to  $  39.50,  how  many  tickets  of  each  kind  were  sold  ? 

22.  A  man  paid  a  bill  of  $16  in  25-cent  pieces  and  5-cent 
pieces.  If  the  number  of  coins  was  80,  how  many  of  each  kind 
were  there  ? 

23.  A  man  paid  $  1  for  some  apples  at  3  cents  each  and  some 
oranges  at  5  cents  each.  He  sold  ^  of  the  apples  and  \  of  the 
oranges  at  cost  for  34  cents.     How  many  of  each  did  he  buy  ? 

24.  A  and  B  together  can  do  a  piece  of  work  in  12  days. 
After  A  has  worked  alone  for  5  days,  B  finishes  the  work  in  26 
days.     In  what  time  can  each  alone  do  the  work  ? 

25.  A  blacksmith  and  his  son  had  a  contract  to  make  a  certain 
number  of  horseshoes.  If  both  had  worked  together,  they  could 
have  done  the  work  in  6  days.  But  the  father  worked  8  days, 
and  the  son  finished  the  work  in  3  days.  In  how  many  days 
could  each  have  done  the  work  ? 

26.  A  man  and  his  two  sons  can  dig  a  ditch  in  6  days ;  if  the 
man  and  either  son  work  7  days,  the  other  son  can  complete  the 
ditch  by  working  2  days.  In  what  time  can  each  alone  dig 
the  ditch? 

27.  A  certain  number  of  persons  agree  to  share  equally  the 
expense  of  hiring  a  coach.  If  each  paid  75  cents,  there  would  be 
$  1.25  over;  but  if  each  paid  50  cents,  there  would  be  $ 2.50  lack- 
ing. What  is  the  number  of  persons  and  the  expense  of  hiring 
the  coach  ? 

28.  A  train  ran  a  certain  distance  at  a  uniform  rate.  Had  the 
rate  been  increased  5  miles  an  hour,  the  journey  would  have  been 
2  hours  shorter;  but  had  the  rate  been  diminished  5  miles  an 
hour,  the  journey  would  have  been  2\  hours  longer.  What  was 
the  distance  and  the  rate  of  the  train  ? 

Suggestion.  —  Let  x  miles  per  hour  be  the  actual  rate  of  the  train  and  y 
the  number  of  hours  required  to  complete  the  journey. 

29.  A  sum  of  money  was  divided  equally  among  a  certain 
number  of  persons.  If  there  had  been  4  persons  more,  the 
share  of  each  would  have  been  $  3  less ;  but  if  there  had  been 


202  SIMULTANEOUS  SIMPLE  EQUATIONS 

2  persons  less,  the  share  of  each  would  have  been  $2  more. 
Among  how  many  persons  was  the  money  divided  and  what  was 
the  share  of  each  ? 

30.  A  dealer  had  eggs  to  sell  and  wished  to  buy  potatoes. 
He  found  that  6  dozen  eggs  were  worth  10  cents  moie  than  2 
bushels  of  potatoes ;  and  that  10  dozen  eggs  were  worth  10  cents 
less  than  4  bushels  of  potatoes.  How  much  were  eggs  and 
potatoes  worth  ? 

31.  If  a  rectangular  floor  were  2  feet  wider  and  5  feet  longer, 
its  area  would  be  140  square  feet  greater ;  if  it  were  7  feet  wider 
and  10  feet  longer,  its  area  would  be  390  square  feet  greater. 
What  are  its  dimensions  ? 

32.  If  54  is  added  to  a  certain  number,  expressed  by  two 
digits  whose  sum  is  8,  the  order  of  the  digits  will  be  reversed. 
What  is  the  niimber  ? 

33.  If  13  is  added  to  a  certain  number,  the  sum  will  be  equal 
to  5  times  the  sum  of  the  two  digits  of  the  number ;  and  if  36  is 
added  to  the  number,  the  order  of  its  digits  will  be  reversed. 
What  is  the  number  ? 

34.  A  and  B  can  do  a  piece  of  work  in  a  days,  or  if  A  works 
m  days  alone,  B  can  finish  the  work  by  working  n  days.  In  how 
many  days  can  each  do  the  work  ? 

35.  A  can  build  a  wall  in  c  days,  and  B  can  build  it  in  d  days. 
How  many  days  must  each  work  so  that,  after  A  has  done  a  part 
of  the  work,  B  can  take  his  place  and  finish  the  wall  in  a  days 
from  the  time  A  began  ? 

36.  One  cask  contains  a  mixture  of  20  gallons  of  wine  and 
30  gallons  of  water,  and  another  contains  a  mixture  of  12  gallons 
of  wine  and  15  gallons  of  water.  How  many  gallons  must  be 
drawn  from  each  cask  to  form  a  mixture  that  will  contain  8  gallons 
of  wine  and  11  gallons  of  water  ? 

Suggestion.  —  If  x  gallons  of  the  mixture  are  drawn  from  the  first  cask, 
I X  gallons  of  it  will  be  wine. 

If  y  gallons  of  the  mixture  are  drawn  from  the  second  cask,  |  y  gallons  of 
it  will  be  wine. 


SIMULTANEOUS   SIMPLE  EQUATIONS  203 

37.  "If  I  had  received  3  oranges  more  for  my  money,"  said 
A,  "  they  would  have  cost  me  1  cent  less  each ;  but  if  I  had 
received  2  less,  they  would  have  cost  me  1  cent  more  each." 
How  many  oranges  had  he  bought,  and  at  what  price  each  ? 

38.  A  merchant  mixes  two  kinds  of  tea.  If  he  mixes  it  in 
parts  proportional  to  7  and  5,  the  value  of  the  mixture  is  46  cents 
a  pound.  If  he  mixes  it  in  parts  proportional  to  5  and  7,  the 
value  of  the  mixture  is  50  cents  a  pound.  What  is  each  kind  of 
tea  worth  per  pound  ? 

39.  A  man  invested  $4000,  a  part  at  5%  and  the  rest  at  4%. 
If  the  annual  income  from  both  investments  was  $  175,  what  was 
the  amount  of  each  investment  ? 

40.  A  man  invested  a  dollars,  a  part  at  ?•%  and  the  rest  at  s% 
yearly.  If  the  annual  income  from  both  investments  was  b 
dollars,  what  was  the  amount  of  each  investment  ? 

41.  A  sum  of  money,  at  simple  interest,  amounted  to  h  dollars 
in  t  years,  and  to  a  dollars  in  s  years.  What  was  the  principal, 
and  what  was  the  rate  of  interest  ? 

42.  A  had  a  certain  sum  invested  at  a  certain  rate  per  cent, 
and  B  had  f  100  less  invested  at  a  rate  2%  higher.  B's  annual 
income  was  $  56  greater  than  A's ;  but  if  B's  rate  upon  his  invest- 
ment had  been  only  1%  higher  than  A's  his  annual  income  would 
have  been  only  $  25  greater  than  A's.  How  much  was  invested 
by  each  man,  and  at  what  rate  ? 

43.  A  crew  can  row  8  miles  downstream  and  back^  Oi- 12  miles 
downstream  and  half  the  way  back  in  1|  hours,  What  is  the 
rate  of  rowing  in  still  water  and  the  velocity  of  the  stream  ? 

44.  A  man  rows  15  miles  downstream  and  oack  in  11  hours. 
The  current  is  such  that  he  can  row  8  miles  downstream  in  the 
same  time  as  3  miles  upstream.  What  is  his  rate  of  rowing  in 
still  water,  and  what  is  the  velocity  of  the  stream  ? 

45.  A  box  will  hold  18  quires  of  paper  and  18  bunches  of 
envelopes,  or  20  quires  of  paper  and  15  bunches  of  envelopes. 
How  many  quires  of  paper  will  the  box  hold  ?  How  many 
bunches  of  envelopes  will  it  hold  ? 


204  SIMULTANEOUS   SIMPLE  EQUATIONS 

46.  A  shelf  will  hold  20  arithmetics  and  24  algebras  oi  14 
arithmetics  and  36  algebras.  How  many  arithmetics  will  the 
shelf  hold  ?     How  many  algebras  will  it  hold  ? 

47.  Two  men  had  a  certain  distance  to  row  and  took  turns  in 
rowing.  Whenever  the  first  rowed,  the  boat  moved  at  a  rate  suf- 
ficient to  cover  the  entire  distance  in  10  hours,  and  whenever  the 
second  rowed,  in  14  hours.  If  the  journey  was  completed  in  12 
hours,  how  many  hours  did  each  row  ? 

48.  A  train  ran  1  hour  and  36  minutes,  and  was  then  detained 
40  minutes.  It  then  proceeded  at  f  of  its  former  rate  and  reached 
its  destination  16  minutes  late.  If  the  detention  had  occurred  10 
miles  farther  on,  the  train  would  have  arrived  20  minutes  late. 
At  what  rate  did  the  train  set  out,  and  what  was  the  whole  dis- 
tance traveled  ? 

49.  A  certain  number  of  people  charter  an  excursion  boat, 
agreeing  to  share  the  expense  equally.  If  each  pays  a  cents, 
there  will  be  h  cents  lacking  from  the  necessary  amount ;  and  if 
each  pays  c  cents,  d  cents  too  much  will  be  collected.  How  many 
persons  are  there,  and  how  much  should  each  pay  ? 

50.  A  sum  of  money  was  to  be  divided  equally  among  a  certain 
number  of  persons,  but  a  persons  more  than  were  expected  ap- 
peared to  claim  a  share,  and  in  consequence  each  received  h  dol- 
lars less.  If  there  had  been  c  persons  less  than  were  expected, 
each  would  have  received  d  dollars  more.  How  many  persons 
were  there,  and  how  much  did  each  receive  ? 

Give  the  results  when  a  =  5,  h  —  100,  c  =  4,  and  d  =  125. 

51.  A  and  B  working  together  can  do  a  piece  of  work  in  a 
days.  But  finding  it  impossible  to  work  at  the  same  time,  A 
works  h  days,  and  later  B  finishes  the  work  in  c  days.  In  how 
many  days  can  each  do  the  work  alone  ? 

If  a  =  5y^y,  6  =  5,  and  c  =  6,  in  how  many  days  can  each  do 
the  work  alone  ? 

52.  A  purse  holds  c  crowns  and  d  guineas;   a  crowns  and  6 

guineas  will  fill  —  th  of  it.     How  many  will  it  hold  of  each  ? 
n 

How  many,  if  c  =  12,  d  =  6,  a  =  4,  6  =  6,  wi  =  1,  and  n  =  2'i 


SIMULTANEOUS   SIMPLE  EQUATIONS  205 

53.  A  mine  is  emptied  of  water  by  two  pumps  which  together 
discharge  m  gallons  per  hour.  Both  pumps  can  do  the  work  in  h 
hours,  or  the  larger  can  do  it  in  a  hours.  How  many  gallons  per 
hour  does  each  pump  discharge  ?  What  is  the  discharge  of  each 
per  hour  when  a  =  5,  &  =  4,  and  m  =  1250  ? 

54.  Two  trains  are  scheduled  to  leave  A  and  B,  m  miles  apart, 
at  the  same  time,  and  to  meet  in  h  hours.  If  the  train  that  leaves 
B  is  a  hours  late  and  runs  at  its  customary  rate,  it  will  meet  the 
first  train  in  c  hours.     What  is  the  rate  of  each  train  ? 

What  is  the  rate  of  each,  if  m  =  800,  c  =  9,  a  =  If,  and  6  =  10  ? 

55.  If  a  quarts  of  good  wine  is  mixed  with  h  quarts  of  poorer 
wine,  the  mixture  will  be  worth  c  cents  a  quart ;  if  h  quarts  of 
the  better  wine  is  mixed  with  a  quarts  of  the  poorer,  the  mixture 
will  be  worth  d  cents  a  quart.  What  is  each  kind  of  wine  worth 
per  quart  ?  What  is  each  kind  of  wine  worth  per  quart,  if  a  =  40, 
6  =  20,  c  =  100,  and  rf  =  80  ? 

212.  Discussion  of  the  general  solution  of  a  system  of  two  simul- 
taneous simple  equations  involving  two  unknown  numbers. 

Let  ax  +  by  —  c  (1) 

and  *  a'x  +  b'y  =  c'  (2) 

be  any  two  simultaneous  simple  equations. 

(1)  X  6',  ab'x  +  bb'y  =  b'c  (3) 

(2)  X  b,  a'bx  +  bb'y  =  be'  (4) 

(3)  -  (4),  iab'-  a'b}x  =  b'c  -  be  (5) 

(1)  X  a',  aa'x  +  a'by  =  ca'  (6) 

(2)  X  a,  aa'x  -t-  ab'y  =  c'a (7) 

(7)-(6),  (ab'-a'b)y  =  c'a-ca'  (8) 

By  the  principles  of  equivalence  the  given  system  may  be  replaced  by  the 

equivalent  system  (5,  8),  in  which  (5)  involves  x  alone  and  (8)  involves  y 

alone.     By  §  197,  each  of  these  simple  equations  involving  one  unknown 

number  has  one  and  only  one  root,  which  can  be  found  except  when  the 

common  coefficient  of  x  and  y  is  equal  to  zero.     Hence,  when  ab'—  a'b  is  not 

equal  to  zero,  the  given  system  is  satisfied  by  one  and  only  one  set  of 

values 

^^b^c-bc'    ^j^^    y  ^  C'a  -  ca' 

ab'—  a'b  ab'  —  a'b 

*  In  algebraic  notation  a',  a",  a'",  etc.,  are  read  'a  prime,'  'a  second,' 
'  a  third,'  etc. 


206  SIMULTANEOUS  SIMPLE  EQUATIONS 

If  ah'  —  a'b  =  0,  that  is,  if  ab'=  a'b,  the  first  members  of  (3)  and  (4)  are 
identical,  and  therefore  the  second  members  must  be  equal.  The  same  is 
true  of  equations  (6)  and  (7).  Hence,  (4)  is  only  a  different  form  of  (3), 
and  (7)  is  only  a  different  form  of  (6);  that  is,  (1)  and  (2)  are  not  inde- 
pendent equations. 

But  if  ab'—  a'b  is  not  equal  to  zero,  neither  (3)  and  (4)  nor  (6)  and  (7) 
are  reducible  to  the  same  form ;  that  is,  (1)  and  (2)  are  independent 
equations. 

It  is  evident  from  the  Distributive  Lav?  for  multiplication  that  the  equations 
(3)  and  (4),  and  also  (6)  and  (7),  cannot  be  combined  by  addition  or  sub- 
traction unless  X  and  y  have  the  same  values  in  (2)  as  in  (1);  that  is,  that 
(1)  and  (2)  cannot  be  solved  unless  they  are  simultaneous  equations. 

It  is  evident  that  equations  (5)  and  (8),  and  therefore  (1)  and  (2),  cannot 
be  solved  if  ab' —  a'b  =  0,  since  §  196,  Prin.  2,  the  members  cannot  be  divided 
by  a  knovrn  expression  equal  to  0 ;  and  it  has  been  shown  that  ( 1 )  and  (2) 
are  dependent  or  independent  equations  according  as  ab'  —  a'b  is  or  is  not 
equal  to  zero. 

Hence,  it  follows  that : 

Two  simple  equations  involving  two  unknoicn  numbers  cannot  he  solved 
unless  the  equations  are  simultaneous  and  independent. 

Evei'y  system  of  two  independent  simultaneous  simple  equations  involving 
two  unknown  numbers  can  be  solved,  and  is  satisfied  by  one,  and  only  one, 
set  of  values  of  its  unknown  numbers. 


THREE  OR  MORE  UNKNOWN  NUMBERS 

213.  1.  In  the  equations  x  +  2 y  -\-  z  =  S  and  2x  +  y  —  z  =  1, 
how  may  z  be  eliminated  ? 

2.  If  one  of  the  unknown  numbers  in  the  above  equations  is 
eliminated,  how  many  unknown  numbers  will  be  left  ? 

3.  How  many  independent  equations  are  necessary  before  the 
values  of  two  unknown  numbers  can  be  found  ? 

4.  How  many  independent  equations  containing  the  same  two 
unknown  numbers  can  be  formed  by  combining  the  equations 
in  (1)  ? 

5.  Since  only  one  derived  equation  containing  two  unknown 
numbers  was  obtained  from  the  given  equations  by  eliminating  z, 
and  since  we  must  have  two  such  equations  before  we  can  find 
the  values  of  x  and  y,  if  another  independent  equation  involving 
aj,  y,  and  z  is  combined  with  either  of  the  equations  in  (1),  how 


SIMULTANEOUS   SIMPLE  EQUATIONS  207 

many  independent  equations  containing  x  and  y  only  will  be 
available  for  finding  the  values  of  x  and  y? 

6.  When  the  values  of  x  and  y  are  found,  how  may  the  value 
of  z  be  found  ? 

7.  Then,  how  many  independent  equations  containing  three 
unknown  numbers  must  be  given,  so  that  the  values  of  the  un- 
known numbers  may  be  found  ?  How  many  to  find  the  values 
of  four  unknown  numbers  ? 

214.  Principle.  —  Every  system  of  independent  simultaneous 
simple  equations  involving  the  same  number  of  unknown  numbers  as 
there  are  equations  can  be  solved,  and  is  satisfied  by  one  and  only 
one  set  of  values  of  its  unknown  numbers. 

The  above  principle  may  be  established  as  follows : 

From  the  given  system  of  n  equations  involving  n  unknown  numbers,  a 
second  system  of  w  —  1  equations  involving  n  —  \  unknown  numbers  may  be 
derived  by  eliminating  one  of  the  unknown  numbers  ;  from  the  second  system 
a  third  system  of  m  —  2  equations  involving  n  —  2  unknown  numbers  may  be 
derived  ;  and  this  process  may  be  continued  until  the  wth  system,  a  single 
simple  equation  involving  only  one  unknown  number,  is  obtained. 

Since,  §  197,  this  equation  has  one  and  only  one  root,  by  substituting  this 
value  in  either  of  the  two  equations  of  the  next  preceding  system  and  solving, 
one  and  only  one  value  of  the  other  number  in  that  equation  is  obtained  ;  by 
substituting  these  two  values  in  any  one  of  the  three  equations  of  the  next 
preceding  system,  one  and  only  one  value  of  the  remaining  unknown  number 
in  that  equation  is  obtained  ;  arid  by  continuing  this  process,  the  value  of 
each  of  the  other  unknown  numbers  is  obtained. 

By  the  principles  of  equivalent  equations,  the  following  system  of  n  equa- 
tions may  be  substituted  for  the  given  system :  the  single  equation  finally 
derived  by  elimination  and  composing  the  nth  system  ;  either  of  the  two 
equations  of  the  preceding,  or  (n  —  l)th  system  ;  any  one  of  the  three  equa- 
tions of  the  system  preceding  that,  or  of  the  (n  —  2)th  system  ;  and  so  on  to 
any  one  of  the  n  equations  in  the  1st  or  given  system. 

But  each  of  the  n  equations  just  described  has  one  and  only  one  value 
of  an  imknown  number.  Hence,  the  given  system  can  be  solved,  and  is 
satisfied  by  one  and  only  one  set  of  values  of  its  unknown  numbers. 

If  the  number  of  unknown  numbers  is  greater  than  the  number  of  inde- 
pendent simultaneous  equations.,  the  last  equation  obtained  by  repeated  elimi- 
nations is  indeterminate,  and  hence  the  system  is  indeterminate. 

If  the  number  of  unknown  numbers  is  less  than  the  number  of  independent 
simultaneous  equations,  say  n  —  p,  any  n  —  p  of  the  equations  involving  the 
n  —  p  unknoion  numbers  form  a  determinate  system. 


208  SIMULTANEOUS  SIMPLE  EQUATIONS 

Examples 

1.    Find  the  values  of  x,  y,  and  2;  in  |  2  a;  +  y  -(-  2  2  =  10, 

I  3 a; +  4?/ -32!  =  2. 

Solution 
a;  +  2y  +  3z  =  14 
2x  +  y  +  2z  =  \0 
33:  +  4y-32  =  2 
Eliminating  z  by  combining  (1)  and  (3), 
(l)  +  (3),  4a:  +  6y  =  16 

Eliminating  z  by  combining  (2)  and  (3), 
(2)x3,  6x  +  3y  +  62  =  30 

(3)x2,  6a;  +  82/-60=4 

Adding,  12  a; +11?/ =  34 

Eliminating  x  by  combining  (7)  and  (4), 
(4)x3,  12a;  +  18y  =  48 

(8) -(7),  ly  =  U 

.-.  2/  =  2 
Substituting  in  (4),  4  a;  +  12  =  16 

.-.  X  =  1 
Substituting  the  values  of  x  and  y  in  (1), 

1  +  4  +  32  =  14  (13) 

.-.2  =  3  (14) 

Explanation. — Eliminating  z  from  (1)  and  (3)  and  from  (2)  and  (3), 
two  simultaneous  equations,  (4)  and  (7),  are  obtained  involving  x  and  y. 
By  the  principle  of  elimination,  §  208,  the  new  systeiii  (1,  4,  7),  or  (2,  4,  7), 
or  (3,  4,  7),  is  equivalent  to  the  given  system. 

Eliminating  x  from  (4)  and  (7),  a  simple  equation  involving  but  one 
unknown  number  y  is  obtained,  and  from  this  equation  the  value  of  y  is 
found,  equation  (10).  Hence,  the  system  (1,  4,  7)  has  been  replaced  by  the 
equivalent  system  (1,  4,  10),  which  is,  therefore,  equivalent  to  the  given 
system. 

Substituting  2  for  y  in  (4),  the  value  of  x  is  found,  giving  a  new  system 
(1,  12,  10)  equivalent  to  the  given  system.  Substituting  the  values  of  both 
X  and  y  in  (1),  the  value  of  ^  is  found,  giving  the  desired  system  (14,  12,  10) 
equivalent  to  the  given  system. 


SIMULTANEOUS   SIMPLE   EQUATIONS 


209 


Solve  the  following : 

r2ic-  y  +  2z  =  12, 
2.  |iB+3?/+  2  =  41, 


3. 


9. 


10. 


11. 


f3a;  +  5y  —  2  =  8, 

I  4a; +  32/ +  22  =  47, 

'  03  +  3  ?/  —  z  =  10, 
2a;  +  5y  +  42!  =  57, 
3a;-2/  +  2z!  =  15. 

a;  +  2/  +  z  =  53, 
a;  +  22/  +  3z  =  105, 
a;  +  32/  +  4z;  =  134. 

a;  -  2/  +  z  =  30, 
3  y  -  a;  -  2  =  12, 
qz  —  y  +  2x=  141. 

|'8a;-52/  +  2z  =  53, 
J  a;  +  y  -  z  =  9, 
[l3a;-9y  +  3z  =  71. 

ra;  +  32/  +  4z  =  83, 
I  a;  +  2/  +  z  =  29, 
l6a;  +  82/  +  3z  =  156. 

2x  +  3?/  +  4z  =  29, 

3a;  +  22/+5z  =  32, 

.4a;  +  32/  +  2z  =  25. 

f  2a; -3?/ +  42-^  =  4, 
4a;  +  2?/  —  z  +  2'u  =  13, 
a;-2/  +  2z  +  3v  =  17, 
3aj  +  22/-2  +  4y  =  20. 

ALG.  — 14 


12. 


13. 


14. 


15. 


16. 


17. 


18. 


3a;-22/  +  z  =  2, 
2x^5y  +  2z  =  21, 
a;  +  3?/  +  3z  =  25. 

4a;-52/  +  3z  =  14, 
X  +  7  2/  —  z  =  13, 
2a;  +  52/  +  5z  =  36. 

f2a;  +  y-3z  +  4w  =  44, 
3  a;  — 2  2/  + z  — «;  =  —  !, 
4a;  —  2/  +  2z  +  w  =  55, 
5a;  —  32/  +  4z  —  w  =  39. 

7a; -1  =  32/, 
11  2  _  1  =  7  ?;, 

]4z-l  =  7?/, 
19a;_l=3u 

\x  +  \y  +  \z=^12. 

ria;-i2/  +  iz  =  3, 
ia;-i2/  +  iz  =  l, 
ia;-i2/  +  iz  =  5. 

■^Jhl  +  3z  =  29, 
3 

-^^-?/  +  2z  =  22, 
2 

3a;-2/  =  3(z-l). 

3a;+?/-z  +  2v  =  0, 
3?/  —  2a;  +  z  —  47;  =  21, 
a;  —  2/  +  2z  —  3-^  =  6, 
4  a;  +  2  2/ -  3  z  +  V  =  12. 


210 


SIMULTANEOUS  SIMPLE   EQUATIONS 


{ u-\-v-\-x-{-y—z=5, 
u+v+x—y  \-z=l, 
19.    Solve  \  u-\-v—x-{-y^z=9, 

u—v+x+y-\-z=H, 
,  v—u-\-x-\-y-\-z=13. 
Solution.  —  Adding  the  equations, 

Dividing  by  3,  u +  v  ->r  x -\-  y  +  z  =  \b. 

Subtracting  each  of  the  given  equations  from  this  equation, 
22-  =  10,    2!/  =  8,    2x  =  6,    2»  =  4,    2w  =  2; 
.  •.  «  =  5,         S/  =  4,       X  =  3,       V  —  2,       M  =  1. 


Solve  the  following : 

f  a;  +  ?/  =  9, 

2^+2  =  7, 

z  -\-  x  =  5. 

■v  +  x  +  y=15, 
x-\-y  +  z  =  18, 

y  +  Z  +  V  =  17, 

2  +  -y  +  a;  =  16. 


20 


21.    ^ 


24. 


25. 


22. 


6, 
10, 


23. 


X     y 
y      z 

Z        X 


_xy_^l 
x-\-y     5' 

yz    ^1^ 
y  +  z     6' 


26. 


27. 


«» 


z  +  x      7 
Suggestion.  —  If 


(  X +  3 y-\-z  =  14:, 
x  +  y  +  3z=16, 
3x  +  y-\-z  =  20. 

'  y  +  z  -\-v  —  x  =  22, 
z  -\-  V  +  X  —  y  =  18, 
V  -^x  +  y  —  z  =  14, 
x-\-y  +  z  —  v  =  10. 

1+^-1  =  0, 

X     y 

1  +  1  +  3  =  0, 

y      z 

-  +  --2  =  0. 

z      X 

_xy_^l^ 

x  +  y     a 

yz   ^1^ 

2/  +  2      b 

zx    _1 

z  +  X      c 


xy 
x  +  y' 


1    x+y     5       ,  lie 

1 775 =  zr ;  whence,  -  +  -  =  6. 

5      xy        1  y     X 


SIMULTANEOUS  SIMPLE  EQUATIONS 


211 


X 

b 

y 

z 

a, 

28.    Solve 

hzx 

— 

cxy  4-  ayz  —  hxyz, 

a 

h 

c  _ 

c. 

X 

y 

z 

Solution 

(l)  +  (3), 

X 

.■.x=   2«. 

a  +  c 

(2)  ^  xyz. 

b_ 

y 

-5  +  «  =  6. 

Z       X 

(5) -(3), 

2fe      ».      ^ 
=0  —  c. 

26 


b-c 
Substituting  the  values  of  x  and  y  in  (1),  and  solving, 

2c 


(1) 
(2) 
(3) 


(4) 
(6) 


29. 


30. 


31. 


32. 


axy  —  x  —  y  =  0, 

bzx  —  z  —  x  =  0, 

.  cyz  —  y  —  z  =  0. 

x  +  y  —  z  =  0, 
X  —  y  =  2  b, 
,x  +  z  =  3a  +  b. 

'  V  -\-  X  =  2  a, 
x-{-y  =  2a  —  z, 
y  +  z  =  a  +  b, 
V  —  z  =  a  -\-  c. 

'y  +  z  —  3x  =  2a, 
z-\-x  —  Sy  =  2b, 
x  +  y  —  3z  =  2c, 

[2x  +  2y-\-v  =  0. 


f  abxyz  -\-  cxy — ayz — bzx = 0, 
33.   ^bcxyz+ayz—bzx—cxy=0, 

[  caxyz  +  bzx — cxy — ayz = 0. 


ix  +  y  +  z  =  a  +  b  +  c, 
b-\-2c, 
&  +  3c. 

'v-{-x  +  y  =  a  +  2b-{-c, 
x  +  y  +  z  =  3b, 
y  +  z+v  =  a  +  b, 

,z-\-v-\-x  =  a  +  3b  —  c. 

'  ax  +  by  -{-  cz  =  3, 
a  +  b 


34.   i  x-{-2y  +  3z 


36. 


36.    i 


x  +  y  = 
y  +  z  = 


ab 

6  +  c 

be 


212  SIMULTANEOUS  SIMPLE  EQUATIONS 

Peoblems 

215.  1 .  Three  men  bought  grain  at  the  same  prices.  A  paid 
$  4.80  for  2  bushels  of  rye,  3  bushels  of  wheat,  and  4  bushels  of 
oats ;  B  paid  f  6.40  for  3  bushels  of  rye,  5  bushels  of  wheat,  and 

2  bushels  of  oats;  and  C  paid  $5.30  for  2  bushels  of  rye,  4 
bushels  of  wheat,  and  3  bushels  of  oats.  What  was  the  price  of 
each? 

2.  A  dealer  was  asked  his  price  for  10  bushels  of  wheat,  corn, 
and  rye.  He  replied,  "  For  5  of  wheat,  2  of  corn,  and  3  of  rye, 
$  6.60 ;  for  2  of  wheat,  3  of  corn,  and  5  of  rye,  $  5.80 ;  and  for 

3  of  wheat,  5  of  corn,  and  2  of  rye,  f  5.60."  What  prices  had 
he  in  mind  ? 

3.  Divide  90  into  three  parts  such  that  the  sum  of  i  of  the 
first,  \  of  the  second,  and  \  of  the  third  shall  be  30;  and  the 
first  shall  be  twice  the  third  diminished  by  twice  the  second. 

4.  There  are  three  numbers  such  that  the  sum  of  \  of  the 
first,  \  of  the  second,  and  \  of  the  third  is  12 ;  of  ^  of  the  first, 
\  of  the  second,  and  \  of  the  third  is  9,  and  the  sum  of  the  num- 
bers is  38.     What  are -the  numbers  ? 

5.  There  are  three  numbers  whose  sum  is  72.  If  the  sum  of 
the  first  two  is  divided  by  the  third,  the  quotient  is  If ;  and  if 
the  third  is  subtracted  from  twice  the  first,  the  remainder  will  be 
\  of  the  second.     Find  the  numbers. 

6.  A  and  B  can  do  a  piece  of  work  in  10  days ;  A  and  C  can 
do  it  in  8  days ;  and  B  and  C  can  do  it  in  12  days.  How  long 
will  it  take  each  to  do  it  alone  ? 

7.  A  certain  number  is  expressed  by  three  digits  whose  sum  is 
14.  If  693  is  added  to  the  number,  the  digits  will  appear  in 
reverse  order.  If  the  units'  digit  is  equal  to  the  tens'  digit 
increased  by  6,  what  is  the  number  ? 

8.  The  third  digit  of  a  number  of  three  digits  is  as  much  larger 
than  the  second  digit  as  the  second  is  larger  than  the  first.  If 
the  number  is  divided  by  the  sum  of  its  digits,  the  quotient  is  15. 
What  is  the  number,  if  the  order  of  its  digits  may  be  reversed  by 
adding  396  ? 


SIMULTANEOUS  SIMPLE  EQUATIONS  213 

9.  Find  three  numbers  such  that  the  first  increased  by  \  of 
the  sum  of  the  other  two  shall  be  36 ;  the  second  increased  by  ^ 
of  the  sum  of  the  other  two  shall  be  40 ;  and  the  third  increased 
by  \  of  the  sum  of  the  other  two  shall  be  44. 

10.  Divide  800  into  three  parts  such  that  the  sum  of  the  first, 
\  of  the  second,  and  f  of  the  third  shall  be  400 ;  and  the  sum  of 
the  second,  |  of  the  first,  and  \  of  the  third  shall  be  400. 

11.  Three  cities,  A,  B,  and  C,  connected  by  straight  roads,  are 
situated  at  the  vertices  of  a  triangle.  From  A  to  B  by  Avay  of  C 
is  130  miles ;  from  B  to  C  by  way  of  A  is  110  miles ;  and  from  C 
to  A  by  way  of  B  is  140  miles.     How  far  apart  are  the  cities  ? 

12.  Find  three  numbers  such  that  the  first  with  \  of  the  sum  of 
the  second  and  third  is  340  ;  the  second  with  \  of  the  sum  of  the 
first  and  third  is  600;  and  the  third  with  \  of  the  remainder 
when  the  first  is  subtracted  from  the  second  is  450. 

13.  A  merchant  has  three  kinds  of  tea.  He  can  sell  2  pounds  of 
the  first  kind,  3  of  the  second,  and  4  of  the  third  for  $  4.70 ;  or  4 
of  the  first,  3  of  the  second,  and  2  of  the  third  for  f  4.30.  If  a 
pound  of  the  third  kind  is  worth  5  cents  more  than  f  of  a  pound 
of  the  first  kind  and  |^  of  a  pound  of  the  second  kind,  what  is  the 
value  of  1  pound  of  each  kind  ? 

14.  A,  B,  and  C  have  certain  sums  of  money.  If  A  gives  B 
$  100,  they  will  have  the  same  amount ;  if  A  gives  C  $  100,  C 
will  have  twice  as  much  as  A;  and  if  B  gives  C  $100,  C  will 
have  4  times  as  much  as  B.     What  sum  has  each  ? 

15.  A  quantity  of  water  sufiicient  to  fill  three  jars  of  different 
sizes  will  fill  the  smallest  jar  4  times ;  the  largest  jar  twice  with 
4  gallons  to  spare ;  or  the  second  jar  3  times  with  2  gallons  to 
spare.     What  is  the  capacity  of  each  jar  ? 

16.  A  gave  to  B  and  C  as  much  as  each  of  them  had ;  B  then 
gave  to  A  and  C  as  much  as  each  of  them  had ;  and  C  then  gave 
to  A  and  B  as  much  as  each  of  them  had,  after  which  each  had 
$  8.     How  much  had  each  at  first  ? 

17.  Three  boys.  A,  B,  and  C,  each  had  a  bag  of  nuts.  After 
each  boy  had  given  each  of  the  others  \  of  the  nuts  in  his  bag, 
they  counted  and  found  that  A  had  740,  B  580,  and  C  380.  How 
many  had  each  at  first  ? 


INVOLUTION 


216.  1.  How  many  times  is  a  number  used  as  a  factor  in  pro- 
ducing its  second  power  ?  its  third  power  ?  its  fourth  power  ?  its 
fifth  power  ?  its  /ith  power,  when  n  is  a  positive  integer  ? 

2.  What  is  the  meaning  of  2^?  of  (-2)3?  of  a'?  oi  (axf? 
of  cc",  when  n  is  a  positive  integer  ? 

3.  What  sign  has  (+  ay?  (+  of  ?  (+  a)",  or  a7iy  power  of  a ? 
What  sign  has  any  power  of  a  positive  number  ? 

4.  What  sign  has  (-a)2?  (-af?  (-a)*?  (-«)«? 

What  sign  have  the  even  powers  of  a  negative  number  ?  What 
sign  have  the  odd  powers  ? 

5.  What  is  the  fourth  power  of  a^  ?  of  a?  ?  of  a^"  ?  of  a",  when 
w  is  a  positive  integer  ?  What  are  the  fifth  powers  of  these  num- 
bers ?  the  sixth  powers  ?  the  mth  powers,  when  m  is  a  positive 
integer  ? 

6.  How  does  8^  compare  in  value  with  2^x4^?  with  22x2^x22? 
32  with  6-  -=-  2^  ?  5=^  with  10=^  -=-22? 

217.  The  process  of  finding  any  required  power  of  an  ex- 
pression is  called  Involution. 

218.  Principles.  — 1.  Law  of  Signs. — All  powers  of  a  positive 
number  are  positive;  even  powers  of  a  negative  number  are  positive, 
and  odd  powers  are  negative. 

2.  Law  of  Exponents. — The  exponent  of  a  power  of  a  number  is 
equal  to  the  exponent  of  the  number  multiplied  by  the  exponent  of  the 
power  to  which  the  number  is  to  be  raised. 

3.  Any  p)oiver  of  a  product  is  equal  to  the  product  of  its  factors 
each  raised  to  that  poiver. 

4.  Any  power  of  the  quotient  of  two  numbers  is  equal  to  the  quo- 
tient of  the  numbers  each  raised  to  that  power. 

214 


INVOLUTION  215 

The  above  principles  may  be  established  as  follows : 
Principle  1  follows  directly  from  the  law  of  signs  for  multiplication. 
Principle  2.     When  m  and  n  are  positive  integers, 
§  24,  (a*")"  =  fl"  X  a"  X  a™  •••  to  n  factors 

;-;  0m+in+m+—  to  n  terms 

=  a™". 
Principle  3.     When  n  is  a  positive  integer, 
§  24,  (o6)"=  ab  X  ab  X  ab  •••  to  n  factors 

§  83,  =(aaa  •••  to  n  factors)  {bbb  •••  to  n  factors) 

=  a»6». 
Principle  4.     When  n  is  a  positive  integer, 

«  .,  « 


§  24,  (^)"  =  -  X  r  X  "  -  to  n  factors 

§180, 


6      b 
aaa  •••  to  n  factors 
bbb  •••  to  n  factors 


6"       - 
219.   Involution  of  monomials. 

Examples 

1.  What  is  the  third  power  of  4  a^h  ? 
Solution.         (4  a^b)^  =  4  a^b  x  4  a^b  x  4  a^b  =  64  a'6'. 

2.  What  is  the  fifth  power  of  -  2  aft^  ? 

Solution.     (-  2o62)5  =  -  2a?;2  x  -2ab^  x  -  2o&2  x  -  2«62  x  -  206' 
=  -  32  asfti". 

To  raise  an  integral  term  to  any  power : 

Rule.  —  Raise  the  numerical  coefficient  to  the  required  power  and 
annex  to  it  each  letter  with  an  exponent  equal  to  the  product  of  its 
exponent  by  the  exponent  of  the  required  power. 

Prefix  the  sign  +  to  any  power  of  a  positive  number  or  to  an  even 
power  of  a  negative  number;  the  sign  —  to  an  odd  power  of  a  nega- 
tive number. 


216 


INVOLUTION 


To  raise  a  fraction  to  any  power : 

Rule.  —  Raise  both  numerator  and  denominator  to  the  required, 
power  and  prefix  the  proper  sign  to  the  result. 

Raise  to  the  power  indicated : 


3. 

(a6V)2. 

4. 

[a^hy. 

5. 

{2a'cf. 

6. 

[1  ahny. 

7. 

{-ly. 

8. 

{-aby. 

9. 

{-3cy. 

10. 

(-10ar^» 

11. 

{-Qa'a^y. 

12. 

(-4cy)^ 

13. 

{-2l*m'dy. 

14. 

{-a^afy"-y. 

27. 

What  is  the  square  of  — 

15. 

[abcxy. 

16.    ( 

;2eV)«. 

17.    ( 

;3  bey. 

18.    ( 

'2  aV)». 

19.    ( 

-^T- 

20.    ( 

-  If. 

21.    ( 

-l)^ 

22.    ( 

_  5)2"+l. 

23.    ( 

^_52^)2„+l_ 

24.    ( 

[—  a^'b^'c"  Hy. 

25.    ( 

'—  a2ny3p^4ry^ 

26.    ( 

[-  a"^'b"-'cf. 

5a^a^, 
7b^c  ' 
Solution 


/     6a%2\2^ 
V      7  b-^c  I 


"b'^c 


7  62c 


=  + 


Raise  to  the  power  indicated ; 


28. 


{M 


33. 


ab 


29.      —- 


30. 


31. 


32. 


2x 

'2xy 

Syj' 

2  6" 


34.       -^ 


35.       -^ 


36. 


37. 


2 
3x^ 

3^Y 

2yj 

_2ay 
x^y) 


25a6x* 
49  6*c2' 


38. 


39. 


40. 


41. 


42. 


_2_a 
b 

bW 
a^xj 

xy*^ 

r,n— 1  /^n\  n 


INVOLUTION  217 

220.  Involution  of  polynomials. 

§  91,  (a  +  by  =  (i^  +  2ah  +  b^ 

§  93,  (a  -  6)2  =  a2  -  2  a6  +  b^. 

§  95,  (a  +  6  +  c)2  =  a2  +  62  +  c2  +  2  a6  +  2  «c  +  2  6c. 

Raise  the  following  to  the  second  power : 

1.  2a +  6.                    5.   3x  —  4.f.  9.  a—h  +  x  —  y. 

2.  2a-b.                    6.    5m*  — 11.  10.  a"  +  a;"  —  2/"+^ 

3.  a^-Sft".                7.   l  —  3abc.  11.  2a  +  3&  — 4c. 

4.  a^  — 2ar^.                8.   4a^  +  5.  12.  5a2— l-f-4n'. 

Raise  to  the  required  power  by  multiplication : 

13.  (x  +  yf.  15.    (x  +  yy.  17.    (x  +  yy. 

14.  (x  —  yy.  16.    (x  —  yy.  18.    {x  —  yy. 

221.  Involution  of  binomials  by  the  Binomial  Theorem. 
By  multiplication, 

(a  +  xy=a?  +  Sa^x  +  Saa^  +  0^. 
(a  -  xy=  a?-3o?x  +  3a2i?  +  af». 
(a  +  a;)*  =  a*  +  4  o?x  +  6  a^'ar'  +  4  aa^  +  a^. 
(a  —  cb)*  =  a"*  —  4  a^a;  +  6  a^x^  —  4  a^  +  a:*, 
(a  +  a;)^  =  a''  +  5  a*x  +  10  a'^a^  +  10  a?a?  +  5  aa;<  +  a^. 
(a  -  a;/  =  a^  -  5  a^o;  +  10  a^'a;-  -  10  a?a?  +  5ax^-  a^. 
Examine  carefully  the  above  powers  of  (a  +  a;)  and  (a  —  x). 

1.  How  does  the  number  of  terms  in  a  power  of  a  binomial 
compare  with  the  exponent  of  the  binomial  ? 

2.  What  terms  of  the  power  contain  the  first  term  of  the  bino- 
mial ?  the  second  term  of  the  binomial  ?  both  terms  ? 

3.  What  is  the  exponent  of  the  first  term  of  the  binomial  in 
the  first  term  of  the  power  ?  in  the  second  ?  in  the  third,  etc.  ? 


218  INVOLUTION 

4.  What  is  the  exponent  of  the  second  term  of  the  binomial  in 
the  second  term  of  the  power  ?  in  the  third  ?  in  the  fourth,  etc.  ? 

5.  What  is  the  coefficient  of  the  first  term  of  the  power? 
How  does  the  coefficient  of  the  second  term  compare  with  the 
exponent  of  the  binomial  ? 

6.  If  the  coefficient  of  any  term  is  multiplied  by  the  exponent 
of  the  first  term  of  the  binomial  found  in  that  term,  and  the 
product  is  divided  by  the  number  oft  the  term,  how  does  the 
quotient  compare  with  the  coefficient  of  the  succeeding  term  ? 

7.  What  are  the  signs  of  the  terms  in  any  power  of  (a  +  &)  ? 
What  terms  are  negative  in  any  power  of  (a  —  b)? 

222.  Principles.  —  1.  The  number  of  terms  in  a  positive  in- 
tegral power  of  a  binomial  is  one  greater  than  the  index  of  the 
required  power. 

2.  Tlie  first  term  of  the  power  contains  only  the  first  term  of  the 
binomial;  the  last  term  of  the  power,  only  the  second  term  of  the 
binomial;  all  other  terms  of  the  power  contain  as  factors  both  terms 
of  the  binomial. 

3.  The  exponent  of  the  first  term  of  the  binomial  in  the  first  term 
of  the  poiver  is  the  same  as  the  index  of  the  required  power,  and  it 
decreases  1  in  each  succeeding  term.  The  exponent  of  the  second 
term  of  the  binomial  in  the  second  term  of  the  power  is  1,  and  it 
increases  1  in  each  succeeding  term. 

4.  Tlie  coefficient  of  the  first  term  of  the  jwwer  is  1.  The  co- 
efficient of  the  second  term  is  the  same  as  the  index  of  the  required 
poiver. 

5.  Tlie  coefficient  of  any  term  may  be  found  by  multiplying  the 
coefficient  of  the  preceding  term  by  the  exponent  of  the  first  term  of 
the  binomial  found  in  that  term,  and  then  dividing  the  result  by  the 
number  of  the  term. 

6.  If  both  terms  of  the  binomial  are  positive,  all  the  terms  of  any 
power  of  the  binomial  will  be  positive. 

7.  If  the  second  term  of  the  binomial  is  negative  and  the  first  term 
positive,  the  terms  of  any  power  of  the  binomial  will  be  alternately 
positive  and  negative. 


INVOLUTION 


219 


Examples 
1.   Find  the  fifth  power  of  (&  —  y)  by  the  binomial  theorem. 

Solution 

Letters  and  exponents,        b^        b*y  b^y^  b^y^        by*     j/* 

Coefficients,  15  10  10  5         1 

Signs,  +         _  +  _  +  _ 


Combined, 


65-5  b*y  +  10  6%2  _  jq  b'Y  +  5by*-y^ 


In  every  term  of  a  power  of  a  binomial  the  sum  of  the  exponents  of  the 
terms  of  the  binomial  is  equal  to  the  index  of  the  required  power. 

Expand : 


2. 

(x  +  yy. 

13. 

(c  -  ny. 

3. 

(m  +  n)*. 

14. 

{x  -  ay. 

4. 

(m  —  ny. 

15. 

{d-yy. 

5. 

(a -0)3. 

16. 

{b  +  yy. 

6. 

(a  +  by. 

17. 

(m  +  ny. 

7. 

(b  +  ay. 

18. 

(p-qy- 

8. 

(q-ry. 

19. 

{s  +  ty. 

9. 

(c  +  ay. 

20. 

ix  +  2y. 

10. 

(x-^yy. 

21. 

{a  +  sy. 

11. 

(x  -  yy. 

22. 

(a; +  4)3. 

12. 

{^-yy- 

23. 

(x  +  5y. 

35. 

Expand  {2h'^ - 

S 

OLUTION 

Let 

2  62  =  a 

,  and  Sy 

Then,                       2  b^ 

-^y  =  a 

-X, 

and 

(262- 

SyY^O 

2-xY 

24. 

[x-2y. 

25. 

{x  +  ry. 

26. 

[b-cy. 

27. 

[p  +  qy. 

28. 

[a  -  by. 

29. 

{a  +  6c)*. 

30. 

[ab  -  cy. 

31. 

(m  —  pny. 

32. 

[m  —  any. 

33. 

[ax  —  byy. 

34. 

[ax  -  byy. 

z=a*  -4a^x  +  6  a^x'^  -  4  aaj'  +  a^ 
Restoring  values,  =  (2  6-^)*  -  4  (2  62)3(3  y)  +  6  (2  62)2(.3  y)^ 

-4(2  62)(3j/)8  +  (3y)* 
=  16  68  _  96  b^y  +  216  6*1/2  -  216  62y8  4.  gl  yi. 


220  INVOLUTION 

36.    Expand  (1  +  x^f- 

Solution 

(1  +  a;2)3  =  13  +  3(l)2(x2)+ 3(l)(a;2)2 +(a;2)8 


41.  (1-Zx^\  45.  {1-xy. 

42.  (5x^-ab)\  46.  {l-2xf. 

43.  {l  +  a?hy.  47.  (a;-i)«. 

44.  {2ax  —  hf.  48.  (^a;  — iy)*. 


52.    f3a2  +  -Y.  55.    f— -2  a; 


Expand : 

37.    (a;  + 

22/)*. 

38.    (2  a; 

-2/)^- 

39.    (2a; 

-5)3. 

40.    (a;^- 

- 10)«. 

Expand : 

49.    (2  a 

-ij 

50.    i^- 

\y 

"•  c^- 

1)- 

6y  V2a; 

53.   M+3^Y.  66.    (l-a^' 


-•  (i4y-    -  ("+r 

58.  Expand  (a-h-cf. 

Solution 

(a  — 6— c)3=(a  — 6  — c)3,  a  binomial  form. 

(a^-c)3=(a-6)3-3(a-6)2c  +  3(a-6)c2-c3 

=  a3-3a26+3a62-63_3c(a2-2a6  +  ft2)  +  3ac2-3  6c2-c8 
=  a3-3  ffl25  +  3  ah^-h^-Z  d^c+Q  a6c-3  62c+3  ac2-3  6c2-c3. 

59.  Expand  {a  +  h  —  c—  df. 

Suggestion.       (o  +  6  —  c  —  d)3  =  (a  +  6  —  c  +  d)3,  a  binomial  form. 

Expand : 

60.  {a  +  x-y)\  66.  (a  +  2b-^cf. 

61.  (a  —  m  —  «)'.  '67.  {a  +  h  +  x  +  y) 

62.  (a  — a; +  2/)^.                     ,  68.  {a  +  h  —  x  —  y) 

63.  (a  —  a;  —  3/)^.  69.  {a  —  h  +  x  —  y) 

64.  (a  +  a;  +  2y.  70.  (a  -  &  -  a;  + 1/)' 

65.  (a -a; -2)'.  71.  (a-b-x-y) 


EVOLUTION 


223.  1.  Of  what  two  equal  numbers  is  16  the  product  ?  What 
is  the  square  root  of  16  ?  Since  16  is  equal  also  to  —  4  x  —  4, 
what  other  square  root  may  16  have  ?  What  is  the  square  root 
of  25  ?  of  64  ?     What  is  the  fourth  root  of  16  ?  of  81  ? 

2.  What  is  the  sign  of  an  even  root  of  a, positive  number? 

3.  Can  the  square  root  of  —  16  be  found  ?  of  —  25  ?  of  —  64? 
the  fourth  root  of  —  16  ?  of  —  81  ?  Can  an  even  root  of  any 
negative  number  be  found  ? 

4.  What  is  the  cube  root  of  8?  of  27?  of  64?  of  -8? 
of  -  27  ?   of  -  64  ?     What  is  the  fifth  root  of  32  ?  of  -  32  ? 

5.  How  does  the  sign  of  an  odd  root  of  a  number  compare 
with  the  sign  of  the  number  ? 

6.  Since  a^=a^  x  a^  X  a^,  what  power  of  a^  is  a®?  What  is 
the  cube  root  of  a^  ?  of  a^  ?  of  a^^  ?  How  is  the  exponent  of  a  in 
the  cube,  or  third,  root  of  any  power  of  a  found  ?  What  is  the 
fourth  root  of  a8  ?  of  a'^? 

7.  How  is  the  exponent  of  a  root  of  a  power  obtained  from 
the  index  of  the  power  and  the  index  of  the  root  ? 


8.  How  does  V4  x  25  compare  in  value  with  V4  x  V25  ? 
V4^  with  V4  X  V9?  \/8  x  1000  with  -^8  x  ^1000?  In  each 
case  how  does  the  root  of  the  product  compare  in  value  with  the 
product  of  the  roots  of  the  factors  ? 


9.  How  does  Vl00-=-4  compare  in  value  with  Vl00-=-V4? 
VW^9  with  V36  -^  V9  ?  -^64 --8  with  ^/6i  --  ^/8  ?  In  each 
case  how  does  the  root  of  the  quotient  compare  in  value  with  the 
quotient  of  the  roots  of  the  dividend  and  the  divisor  ? 

221 


222  EVOLUTION 

224.  The  process  of  finding  any  required  root  of  an  expression 
is  called  Evolution. 

225.  Since  the  product  of  two  numbers  having  like  signs  is 
positive,  every  positive  number  has  two  square  roots,  numerically 
equal,  hut  with  opposite  signs.  It  will  be  seen  later  that  every 
number  has  two  square  roots,  three  cube  roots,  four  fourth  roots, 
five  fifth  roots,  and,  in  general,  q  gth  roots.  Of  these  roots  the 
positive  roots  of  positive  numbers  and  the  negative  odd  roots  of 
negative  numbers  are  called  Principal  Roots. 

\/25  =  +  5  or  —  5,  and  +  5  is  the  principal  square  root.  \/l6  =  +  2  or 
—.2  or,  as  will  be  seen  later,  +  V— 4  or  —  V—  4,  but  +  2  is  the  principal 
root.    The  principal  cube  root  of  8  is  +  2  and  of  —  8  is  —  2. 

226.  A  number  that  is  or  can  be  expressed  as  an  integer  or  as 
a  fraction  with  integral  terms  is  called  a  Rational  Number. 

a,  3,  5|,  a^  _|_  ^^2^  V25,  and  ..333  are  rational  numbers. 

A  number  that  cannot  be  expressed  as  an  integer  or  as  a  frac- 
tion with  integral  terms  is  called  an  Irrational  Number. 

When  the  indicated  root  of  a  number  cannot  be  exactly  obtained, 
the  root  is  irrational. 

The  indicated  roots  V2,  -y/i,  VoM^,  v^,  \/x^,  and,  in  general,  the 
gth  root  of  a  number  that  is  not  the  gth  power  of  some  rational  number,  are 
irrational  numbers. 

227.  A  rational  arithmetical  number  is  called  a  Commensurable 
Number,  and  an  irrational  arithmetical  number  is  called  an  Incom- 
mensurable Number. 

2,  f,  .54,  and  .666  are  commensurable,  but  V2  is  incommensurable. 

In  algebra,  commensurable  and  incommensurable  numbers  may 
be  either  positive  or  negative. 

The  terms  rational  and  irrational  applied  to  algebraic  numbers 
relate  to  their  forms,  while  the  terms  commensurable  and  incom- 
mensurable relate  to  their  arithmetical  values. 

3  a,  a  +  b,  x  —  3,  x^,  are  rational  but  not  necessarily  commensurable. 
For  a  may  represent  \/2,  b  may  represent  \/5,  etc.  Again,  Vx  is  irrational, 
but  if  X  =  16,  Vx  is  commensurable. 

Incommensurable  numbers  obey  the  Commutative,  Associative,  and  Dis- 
tributive Laws,  but  the  proof  is  too  complicated  to  be  given  here. 


EVOLUTION  223 

228.  Since  the  wth  power  of  a  number  is  the  product  of  n  equal 
factors,  and  since  one  of  these  factors  is  a  root  of  the  power,  it 
follows  that  ( >/«)"= a,  and  if  the  principal  root  is  meant,  -\/a"=a, 
when  11  is  a  positive  integer. 

In  the  statement  of  the  following  principles  and  iri  Ax.  7  the 
term  root  means  principal  root. 

229.  Principles.  —1.  Law  of  Signs.  —  An  odd  root  of  a  number 
has  the  same  sign  as  the  number ;  an  even  root  of  a  jJositive  number 
is  positive  ;  an  even  root  of  a  negative  number  is  impossible,  or 
imaginary. 

2.  Law  of  Exponents.  —  Tlie  exponent  of  any  root  of  a  number  is 
equal  to  the  exponent  of  the  given  number  divided  by  the  index  of  the 
root. 

3.  Any  root  of  a  product  is  equal  to  the  product  of  that  root  of 
each  of  the  factors. 

4.  Any  root  of  the  quotient  of  two  numbers  is  equal  to  the  qiiotient 
of  that  root  of  each  of  the  numbers. 

Even  roots  of  negative  numbers  will  be  discussed  later. 

The  above  principles  may  be  established  as  follows : 

Pbinciple  1  follows  from  the  Law  of  Signs  for  multiplication. 

Principle  2.     When  m  and  n  are  positive  integers, 
§  218,  Prin.  2,  a"*"  =  (a")". 

Taking  the  Jith  root,  §  228,  Va^  =  a^  ; 

.-.  y/a^  =  a"""^  =  a™. 

Thus,  \^aP  =  a2. 

Principle  3.     When  n  is  a  positive  integer, 
§  228,  §  218,  Prin.  3,  ab  =  (  y/a)«  x  (  v^)»  =  (  7a  x  V6)» 

Taking  the  nth  root,  Vab  =  Va  x  Vb. 

Thus,  v/27a9  =  ^/2^  x  v^  =  3 a«. 

Principle  4.     When  n  is  a  positive  integer, 

§228,  §218,  Prin.  4,  ?  =  i^"=f-fiy. 

Taking  the  nth  root,  r'k  =  "^. 


^4  EVOLUTION 

230.   Evolution  of  monomials. 

Examples 

1.  What  is  the  square  root  of  36  a^6^  ? 

Solution.  —  Since,  in  squaring  a  monomial,  §  218,  the  coefficient  is 
squared  and  the  exponents  of  the  letters  are  multiplied  by  2,  to  extract  the 
square  root,  the  square  root  of  the  coefficient  must  be  found,  and  to  it  must 
be  annexed  the  letters  each  with  its  exponent  divided  by  2. 

The  square  root  of  36  is  6,  and  the  square  root  of  the  literal  factors  is  a^h. 
Therefore,  the  principal  square  root  of  36  a^h'^  is  6  a^b. 

The  square  root  may  also  be  —  6  a^fe,  since  —  6  a%  x  —  6  a^b  =  36  a^V^. 

.:  V36a6p  =  ±  6 aSft. 

2.  What  is  the  cube  root  of  -  125  a^y^  ? 
Solution.  y/-l25a^y^>-  =  -  5  x^y''. 

To  find  the  root  of  an  integral  term : 

Rule.  —  Extract  the  required  root  of  the  numerical  coefficient, 
annex  to  it  the  letters  each  with  its  exponent  divided  by  the  index  of 
the  root  sought,  and  prefix  the  proper  sign  to  the  residt. 

To  find  the  root  of  a  fractional  term : 

Rule.  —  Find  the  required  root  of  both  numerator  and  denomi- 
nator and  prefix  the  proper  sign  to  the  resulting  fraction. 

Find  the  indicated  root : 


3. 

-^/a^^c"^. 

4. 

Va«6^V*. 

5. 

S/a^V^. 

6. 

A/a^"6«c^. 

7. 

Vic^Yz^". 

8. 

A/-8a«6«. 

9. 

^-32aJY« 

10. 

V16  xy. 

11. 

■</ -  a'^b^x'*. 

12. 

A/-243y*». 

13. 

a/IG  mV. 

14. 


15. 


16. 


17. 


18. 


rl(x-yy* 
\  128  a;" 

5/-32aV» 
\    243  yi« 


8  256  ar8 


6561 


3  /     125  a;  V 
\        1728  c' 


\      ^4r 


2»6* 


2.4ny2n 


'    \2"ar^"^' 


EVOLUTION  225 

231.    To  extract  the  square  root  of  a  polynomial. 

1.  Since   o?  +  2ab  +  b'^  is  the  square  of  (a  +  6),  what  is  the 
■  square  root  of  a^ -{-2iab  +¥? 

2.  How  may  the  first  term  of  the  square  root  be  found  from 
a^  +  2  a6  +  6'  ? 

3.  How  may  the  second  term  of  the  square  root  be  found  from 
2  ah,  the  second  term  of  the  power  ? 

4.  What  are  the  factors  of  o?+2db  +  h'^? 

5.  Since  2ab  +  b^  is  equal  to  b(2a  +  b),  what  are  the  factors 
of  the  last  two  terms  of  the  square  of  a  binomial  ? 

6.  By  what  divisor,  then,  must  the  last  two  terms  of 
a^  -\-  2  ab  -\- b^  be  divided  so  that  the  quotient  may  be  the  second 
term  of  the  square  root  ? 

Examples 

1.  Find  the  process  for  extracting  the  square  root  of 
a''  +  2ab  +  b\ 


PROCESS 

a^  +  2  a6  +  6"  I  a  +  & 


a 


Trial  divisor,  2  a 
Complete  divisor,  2a-\-b 


2ab-{-b^ 
2ab-\-b^ 


Explanation.  — Since  a^  +  2  a6  +  b'^  is  the  square  of  (a  +  6),  we  know 
that  the  square  root  of  a^  -{-  2  ab  +  b^  is  a  +  b. 

Since  the  first  term  of  the  root  is  a,  it  may  be  found  by  taking  the  square 
root  of  a^,  the  first  term  of  the  power.  Subtracting  a^,  there  is  a  remainder 
of  2  ah  +  62. 

The  second  term  of  the  root  is  known  to  be  b,  and  that  may  be  found  by 
dividing  the  first  term  of  the  remainder  by  twice  the  part  of  tlie  root  already 
found.     This  divisor  is  called  a  trial  divisor. 

Since  2ab  +  b^  is  equal  to  6(2  a  +  b),  the  complete  divisor  which  multi- 
plied by  6  produces  the  remainder  2  a6  +  6^  is  2  a  +  6  ;  that  is,  the  complete 
divisor  is  found  by  adding  the  second  term  of  the  root  to  twice  the  root 
already  found. 

Multiplying  the  complete  divisor  by  the  second  term  of  the  root  and  sub- 
tracting, there  is  no  remainder,  consequently,  a  +  6  is  the  required  root. 

ALG.  — 16 


226  EVOLUTION 

2.   Find  the  square  root  of  9oc? -SQxy -^-25^. 

PKOCESS 

9 x2 -  30 xt/  +  25 y2  [  3a;-5y 

9  or 

6a; 


6x  —  5y 


-  30  x?/  +  25  / 

-  30  ic^  +  25  / 


Since,  in  squaring  a  +  h  -\-c,a  +  h  may  be  represented  by  x,  and 
the  square  of  the  number  by  ar^  +  2  a;c  +  c^,  it  is  obvious  that  the 
square  root  of  a  number  whose  root  consists  of  more  than  two 
terms  may  be  extracted  in  the  same  way  as  in  Example  1,  by  con- 
sidering the  terms  already  found  as  one  term. 

3.   Find  the  square  root  of  4  a;*  +  12  a;^  —  3  a.-^  - 18  a;  +  9. 

PROCESS 

4a;<  +  12ar»-    3a;2_i8a;  +  9  |  2a;^  +  3a;-3 

4  a!*     

12x3-    3a^ 


4x2 

4x2  +  3^ 


12x3+    g^ 


4x2  +  6x  |-12x2-18x  +  9 

4x'4-6x-3  I  -12x2-18x  +  9 

Explanation.  —  Proceeding  as  in  the  previous  example,  the  first  two 
terms  of  the  root  are  found  to  be  2  x^  +  3  x. 

Considering  (2  x^  +  3  x)  as  the  first  term  of  the  root,  the  next  term  of  the 
root  is  found  as  the  second  term  of  a  root  is  found,  by  dividing  the  remainder 
by  twice  the  part  of  the  root  already  found.  Hence,  the  trial  divisor  is 
4  x^  +  6  X,  and  the  next  term  of  the  root  is  —  3.  Annexing  this,  as  before, 
to  the  trial  divisor  already  found,  the  entire  divisor  is  2  x'^  +  3  x  —  3.  This 
multiplied  by  —  3  and  the  product  subtracted  from  —  12  x^  —  18  x  +  9,  leaves 
no  remainder.     Hence,  the  square  root  of  the  number  is  2  x'-^  +  3  x  —  3. 

Rule.  —  Arrange  the  terms  of  the  polynomial  with  reference  to 
the  consecutive  powers  of  some  letter. 

Extract  the  square  root  of  the  first  term,  write  the  result  as  the 
first  term  of  the  root,  and  subtract  its  square  from  the  given 
polynomial. 


EVOLUTION  227 

Divide  the  first  term  of  the  remainder  by  twice  the  root  already 
found,  as  a  trial  divisor,  and  the  quotient  will  he  the  next  term  of 
the  root.  Write  this  result  in  the  root,  and  annex  it  to  the  trial 
divisor  to  form  the  complete  divisor. 

Multiply  the  complete  divisor  by  this  term  of  the  root,  and  sub- 
tract the  product  from  the  first  remainder. 

Find  the  next  term  of  the  root  by  dividing  the  first  term  of  the  re- 
mainder by  the  first  term  of  the  trial  divisor. 

Form  the  complete  divisor  as  before  and  continue  in  this  manner 
until  all  the  terms  of  the  root  are  found. 

Find  the  square  root  of  the  following : 

4.  4x2  + 12a; +  9.  7.   3^  +  xy-^\if. 

5.  4  ^2  ^  20  a;  +  25.  8.   4.  x?  -  52  x  +  169. 

6.  25a^  +  40a;  +  16.  9.    (a  +  6)2- 4(a  +  6) +4. 

10.  a;^  +  4a^  +  2a;^  +  9a;2  — 4a;  +  4. 

11.  9  x*  -  12  or' +  10  ar*- 4  a; -f-1. 

12.  x*-6ci^y-\-13x'y^-12xf-\-4:y*. 

13.  a^  +  2  a^x^  -  a'x*  -  2  a^af  +  a\ 

14.  25a;* +  4 -12a;- soar' +  29x2. 

15.  l  +  6x  +  15x'-\-20a^  +  15x*  +  6x^-{-a^. 

16.  l-2x  +  3x^-4:X^-\-3x!' -2x^  +  3^. 

17.  a*-2a^b  +  2a^c^-2b(^  +  b^  +  c*. 

18.  4  a"  -  12  a&  +16  ac  +  9  6^  +  16  c^  -  24  be. 

19.  9a;2_^25/  +  9z2_30a;?/  +  18a;;?-30^2. 

20.  a;2_^2a;-l--  +  -. 

X        XT 

21.  x*  +  a?-\-'^^  +  ^-\-—- 

20        5     25 

22.  -  +  aa;  +  -^+    ^     f- 

23.  a;8  +  4a;'-2a;«-20ar*-3a;*  +  32a;'  +  4a;2_iga._,_4 


228  EVOLUTION 

24.    Find  four  terms  of  the  square  root  of  1  +  x. 

Solution 


14 

X\  l  +  |x-jtx2  +  A** 

1 

2  +  ix 

X 

X+     ix2 

2  +  a;  -  ^  x2 

-    ix2 

2 +  x- 1x2+^^x3  I  1x3 -^V  a;* 
Find  the  square  root  of  the  following  to  four  terms : 

25.  1-a.  27.    ar^-1.  29.    ?/ +  3. 

26.  a?  +  l.  28.    4  -  a.  30.    a^  +  2b. 

SQUARE  ROOT  OP  ARITHMETICAL  NUMBERS 

1-  =  1  10''  =  100  100^  =  10000 

9"  =  81  992  =  9801  9992  =  998001 

232.  1.  How  many  figures  are  required  to  express  the  square 
of  a  number  expressed  by  1  figure  ?  2  figures  ?  3  figures  ?  4 
figure's  ? 

2.  How  does  the  number  of  figures  in  the  square  of  a  number 
compare  with  the  number  of  figures  in  the  number  ? 

3.  How  many  figures  are  there  in  the  square  root  of  a  number 
that  is  expressed  by  4  figures  ?  by  3  figures  ?  by  5  figures  ? 
by  6  figures  ?   by  7  figures  ? 

4.  How,  then,  may  the  number  of  figures  in  the  square  root  of 
a  given  number  be  found  ? 

Principles.  —  1.  The  square  of  a  number  is  expressed  by  twice 
as  many  figures  as  there  are  in  the  number  itself,  or  by  one  less  than 
twice  as  many. 

2.  TJie  orders  of  units  in  the  square  root  of  a  number  correspond 
to  the  number  of  ji^riods  of  two  figures  each  into  ivhich  the  number 
can  be  separated,  beginning  at  units. 


EVOLUTION  229 

233.  If  the  number  of  units  expressed  by  the  tens'  digit  is 
represented  by  t  and  the  number  of  units  expressed  by  the  units' 
digit  by  w,  the  square  of  a  number  consisting  of  tens  and  units 
will  be  represented  by  (t  +  uY,  or  f  ■\-2tu-{-  v?. 

Thus,  25  =  2  tens  +  5  units,  or  20  +  5  units, 

and  252  ^  202  +  2(20  +  5)  +  52  =  625. 

Examples 
1.   What  is  the  square  root  of  2809  ? 

Explanation.  —  Accordinff  to  Prin. 

FIRST    PROCESS  ^      „  „„,       ,  ,  .  .        .         , 

2,    §  231,  the  orders  of  unus   m  the 

28-09  I  50  +  3         square  root  of  a  number  may  be  deter- 

t^  =  25  00  mined  by  separating  the  number  into 

0  ^Q  periods  of  two  figures  each,  beginning 


2«  =  100 

u  =      3 


2t  +  ^l  =  10^ 


at  units.     Separating  2809  thus,  there 
are  found  to  be  two  orders  of  units  in 
the  root ;  that  is,   it  is  composed  of 
o  /-kq  tens  and  units. 

Since  the  square  of  tens  is  hundreds, 


and  the  hundreds  of  the  power  are  less  than  36,  or  6^,  and  more  than  25, 
or  52,  the  tens'  figure  of  the  root  must  be  5.  5  tens,  or  50,  squared  is  2500, 
and  2500  subtracted  from  2809  leaves  309,  which  is  equal  to  2  times  the 
tens  X  the  units  +  the  units2. 

Since  two  times  the  tens  multiplied  by  the  units  is  much  greater  than  the 
square  of  the  units,  309  is  a  little  more  than  2  times  the  tens  multiplied  by 
the  units.  Therefore,  if  309  is  divided  by  2  times  the  tens,  or  100,  the  trial 
divisor,  the  units  are  found  to  be  3.  And  since  the  complete  divisor  is  found 
by  adding  the  units  to  twice  the  tens,  the  complete  divisor  is  100  +  3,  or  103. 
This  multiplied  by  3  gives  as  a  product  309,  which  subtracted  from  309  leaves 
no  remainder.     Therefore,  the  square  root  of  2809  is  53. 

SECOND    PROCESS 

^o«uy  I  OtJ  Explanation. — In  practice  it  is  usual  to 

place  the  figures  of  the  same  order  in  a  col- 
309  umn,  and  to  disregard  the  ciphers  on  the 

right  of  the  products. 


«==  25 

2«=100 
u  =      3 
2t  +  u  =  103 


309 


Since  any  number  may  be  regarded  as  composed  of  tens  and 
units,  the  processes  given  above  have  a  general  application. 
Thus,  346  =  34  tens  +  6  units  ;  2377  =  237  tens  +  7  units. 


230 


EVOLUTION 


2.   Find  the  square  root  of  104976. 


Solution 


Trial  divisor         =  2  x  30  =60 
Complete  divisor  =  60  +  2  =62 
Trial  divisor         =  2  x  320  =  640 
Complete  divisor  =  640  +  4  =  644 


10.49.76 
^ 

1  49 

124 
25  76 
25  76 


324 


Rule.  —  Separate  the  number  into  periods  of  two  figures  each, 
beginning  at  units. 

Find  the  greatest  square  in  the  left-hand  period  and  ivrite  its  root 
for  the  first  figure  of  the  required  root. 

Square  this  root,  subtract  the  result  from  the  left-hand  period,  and 
annex  to  the  remainder  the  next  period  for  a  new  dividend. 

Double  the  root  already  found,  with  a  cipher  annexed,  for  a  trial 
divisor,  and  by  it  divide  the  dividend.  The  quotient  or  the  quotient 
diminished  will  be  the  second  figure  of  the  root.  Add  to  the  trial 
divisor  the  figure  last  found,  midtiply  this  complete  divisor  by  the 
figure  of  the  root  found,  subtract  the  product  from  the  dividend,  and 
to  the  remainder  annex  the  next  period  for  the  next  dividend. 

Proceed  in  this  manner  until  all  the  periods  have  been  used.  The 
result  will  be  the  square  root  sought. 

1.  When  the  number  is  not  a  perfect  square,  annex  periods  of  decimal 
ciphers  and  continue  the  process. 

2.  Decimals  are  pointed  off  from  the  decimal  point  toward  the  right. 

3.  The  square  root  of  a  fraction  may  be  found  by  extracting  the  square 
root  of  both  numerator  and  denominator  separately  or  by  reducing  it  to  a 
decimal  and  then  extracting  its  root. 

Extract  the  square  root  of  the  following : 

3.  529.  9.  57121.  15.  125316. 

4.  2209.  10.  42025.  16.  455625. 

5.  4761.  11.  95481.  17.  992016. 

6.  7921.  12.  186624.  18.  2480.04. 

7.  17424.  13.  165649.  19.  10.9561. 

8.  19321.  14.  134689.  20.  .001225. 


EVOLUTION 


231 


21.  10201. 

22.  95481. 

23.  363609. 
30 
31 


625 


24.  1332.25. 

25.  101.0025. 

26.  111.0916. 
32.  HI-       34 
33-  TT5^*       35 

Extract  the  square  root  to  four  decimal  places : 

38.  f.  40.    |.  42 

39.  f.  41.    .6.  43 


576 


27. 

540.5625. 

28. 

1.018081. 

29. 

13003236. 

Ml- 

36.  m- 

361 
400- 

37.  m- 

places : 

5 

44.    f 

1- 

45.    ^. 

234.   To  extract  the  cube  root  of  a  polynomial. 

1.  Since  a^  +  3  a^6  +  3  ab^  +  &'  is  the  cube  of  (a  +  h),  what  is 
the  cube  root  of  o?  +  3  a^S  +  3  ab^  +  6^  ? 

2.  How  may  the  first  term  of  the  root  be  found  from  a?  +  3  a^h 

3.  How  may  the  second  term  of  the  root  be  found  from  the 
second  term  of  the  power,  3  a?b  ? 

4.  What  are  the  factors  of  3  aj^h  +  SaW  +  W? 

5.  Since  3  a^^fi  +  3  ah'^  +  W  is  equal  to  6  (3  a^  +  3  a&  +  h%  by 
what  number  must  the  last  three  terms  of  a^  +  3  a?h  +  3  a6^  +  If 
be  divided  so  that  the  quotient  may  be  the  second  term  of  the 
cube  root? 

Examples 

1.    Find  the  process  for  extracting  the  cube  root  of  a'-'t^a^h 

PROCESS 

a^  +  3 a?h  +  3 aft^  +  h^  \a  +  b 
a? 

Trial  divisor,  3  a? 

Complete  divisor,   3  a^+  3  a6  +  6^ 


Za?b  +  SaJf+W 


Explanation.  — Since  a^  +  3  a'^6  +  3  aV^  +  h^  is  the  cube  of  (a  +  6),  we 
know  that  the  cube  root  of  a'  +  3  a%  -{■  Z  ah^  +  b^  is  a  +  h. 

Since  the  first  term  of  the  root  is  a,  it  may  be  found  by  taking  the  cube 
root  of  a^,  the  first  tern?  of  the  power.  Subtracting,  there  is  a  remainder  of 
3  a26  +  3  ab^  +  ¥. 


232  EVOLUTION 

The  second  term  of  the  root  is  known  to  be  6,  and  that  may  be  found  by 
dividing  the  first  term  of  the  remainder  by  3  times  the  square  of  tlie  part  of 
the  root  already  found.     This  divisor  is  called  a  trial  divisor. 

Since  3  a%  +  3  ah^  +  b^  is  equal  to  h  (.3  a"^  +  ^ah  +  b^),  the  complete  divi- 
sor, which  multiplied  by  b  produces  the  remainder  3  a^b  +  3  ab^  +  b^,  is 
S  a^  +  Sab  +  b'^ ;  that  is,  the  complete  divisor  is  found  by  adding  to  the 
trial  divisor  3  times  the  product  of  the  first  and  second  terms  of  the  root  and 
the  square  of  the  second  term  of  the  root. 

Multiplying  the  complete  divisor  by  the  second  terra  of  the  root,  and  sub- 
tracting, there  is  no  remainder ;  consequently,  a  +  6  is  the  required  root. 

Since,  in  cubing  a-f-6  +  c,  a  +  b  may  be  expressed  by  x,  its 
cube  will  he  a^  +  3  x'c  +  3  xc'  +  c^.  Hence,  it  is  obvious  that  the 
cube  root  of  an  expression  whose  root  consists  of  more  than  two 
terms  may  be  extracted  in  the  same  way  as  in  example  1,  by  con- 
sidering the  terms  already  found  as  om  term. 

2.    Find  the  cube  root  ot  b^  -  3  b' +  5b^ -3b -1. 

PROCESS 

b^-3b'>+5b^-3b-l[b^~b-l 
^ 

Trial  divisor,  3  b* 

Complete  divisor,    3b*-3b^+b^ 


-3b'+5b^ 
-3b'+3b'-W 


Trial  divisor,  3  &*-6  W-\-3  b^ 


Complete  divisor,    3 b*-6  5^+3 &  + 1    -3  6^-1-6  6^-3  6-1 


-3  6^-^-6  63-3  6-1 


Explanation.  —  The  first  two  terms  are  found  in  the  same  manner  as  in 
the  previous  example.  In  finding  the  next  term,  b^  —  b  is  considered  as  one 
term,  which  we  square  and  multiply  by  3  for  a  trial  divisor.  Dividing  the 
remainder  by  this  trial  divisor,  the  next  term  of  the  root  is  found  to  be  —  1. 
Adding  to  the  trial  divisor  3  times  (b^  —  b)  multiplied  by  —  1,  and  the  square 
of  —  1,  we  obtain  the  complete  divisor.  This  multiplied  by  —  1,  and  the 
product  subtracted  from  —Sb*  +  6b^  —  36  —  1,  leaves  no  remainder.  Hence, 
the  cube  root  of  the  polynomial  is  6^  _  ^  _  i. 

Rule.  —  Arrange  the  polynomial  with  reference  to  the  consecutive 
powers  of  some  letter. 

Extract  the  cube  root  of  the  first  term,  write  the  result  as  the  first 
term  of  the  root,  and  subtract  its  cube  from  the  given  polynomial. 

Divide  the  first  term  of  the  remainder  by  3  times  the  square  of 
the  root  already  found,  as  a  trial  divisor,  and  the  qxiotient  will  be 
the  next  term  of  the  root. 


EVOLUTION  233 

Add  to  this  trial  divisor  3  times  the  product  of  the  first  and  second 
terms  of  the  root,  and  the  square  of  the  second  term.  TJie  result 
will  be  the  complete  divisor. 

Multiply  the  complete  divisor  by  the  last  term  of  the  root  found, 
and  sxibtract  this  product  from  the  dividend. 

Find  the  next  term  of  the  root  by  dividing  the  first  term  of  the 
remainder  by  the  first  term  of  the  trial  divisor. 

Form  the  complete  divisor  as  before,  and  continue  in  this  manner 
until  all  the  terms  of  the  root  are  found. 

Find  the  cube  root  of 

3.  or'  —  3  x^y  -\-  'S  xy^  —  i^. 

4.  m^-9m24-27m-27. 

5.  8  m^  -  60  mhi  +  150  mn^  -  125  n\ 

6.  27  a^  -  189  a;"^/ +  441  0^2/2  -  343  y'. 

7 .  125  a^  +  675  a^x  +  1215  aa?  +  729  ^. 

8.  1000  j9«  -  300  p*q  +  30 p'^q'  -  q\ 

9.  m^  +  6  m^  +  15  m'^  +  20  m^  +  15  m^  4-  6  m  +  1. 

10.  a:^  -  6  ar*  +  15  a;^  -  20  a^  +  15  a^- 6  a; +  1. 

11.  a^  +  3  ar*  +  9  a*  4- 13  a^  +  18  ar^+ 12  a; +  8.   - 

12.  a^  +  12  ar^  +  63  a;*  +  184  \i?  +  315  ar'  +  300  x  +  125. 

13.  a;«  4-  6  a;^  -  18  a;*  -  1000  +  180  a^  -  112  x?  +  600  x. 

14.  a^-12ar^  +  54a;-112  +  ^-^  +  -^. 

a;        ar      ar 

15.  1  -  6  a  +  21  a"  -  44  a^  +  63  a^  -  54  a«  +  27  a\ 

16.  64  -  144p  +  156 p''  -  99i)3  +  39 p*  -  Qp"  +p\ 

^^     a^Wx>  _<M     Zacx^     3  a^bx^ 


c?  b^  b 


c 


18.   a:«+15x2  +  ^  +  20  +  -^  +  ^  +  6a^, 
ar  a;*     ar 


234  EVOLUTION 

CUBE  ROOT  OF  ARITHMETICAL  NUMBERS 

1^  =  1.  10^  =  1000.  100^  =  1000000. 

38  =  27.  303  =  27000.  3003  =  27000000. 

93  =  729.  993  =  970299.  999^  =  997002999. 

235.  1.  How  many  figures  are  there  in  the  cubes  of  numbers 
that  have  1  figure  ?    2  figures  ?    3  figures  ?   4  figures  ? 

2.  What  places,  then,  belong  to  the  cube  of  units  ?  of  tens  ? 
of  hundreds  ?   of  thousands  ? 

3.  How  many  figures  are  there  in  the  cube  root  of  a  number 
expressed  by  4  figures  ?   5  figures  ?   6  figures  ?   7  figures  ? 

4.  How  may  the  number  of  figures  in  the  cube  root  of  a 
number  be  found? 

Principles.  —  1.  Tlie  cube  of  a  number  is  expressed  by  three 
times  as  many  figures  as  the  number  itself,  or  by  one  or  two  less 
than  three  times  as  many. 

2.  The  orders  of  units  in  the  cube  root  of  a  number  correspond  to 
the  number  of  periods  of  three  figures  each,  into  which  the  number 
can  be  separated,  beginning  at  units. 

236.  If  the  number  of  units  expressed  by  the  tens'  digit  is 
represented  by  t,  and  the  number  of  units  expressed  by  the  units' 
digit  by  m,  the  cube  of  a  number  consisting  of  tens  and  units  will 
be  represented  by  (t  +  uf,  or  ^  +  3  t^u  +3tu^  +  u^. 

Thus,  25  =  2  tens  +  5  units,  or  20  +  5  units, 

and  253  =  203  4.  3  (202  x  5)  +  3  (20  x  52)  +  58  =  15625. 

Examples 
1.    What  is  the  cube  root  of  12167  ? 

FIRST    PROCESS 

12-167  [20  +  3 
«3  =  8  000 

Trial  divisor,  Sf  =  1200 

Stu^   180 
m2=       9 


Complete  divisor,  =  1389 


4  167 


4  167 


EVOLUTION 


235 


Explanation.  —  By  Prin.  2,  §  235,  the  orders  of  units  may  be  determined 
by  separating  the  number  into  periods  of  three  figures  each,  beginning  at 
units.  Separating  12167  thus,  there  are  found  to  be  two  orders  of  units  in 
the  root ;  that  is,  the  root  is  composed  of  tens  and  units. 

Since  the  cube  of  tens  is  thousands,  and  the  thousands  of  the  power  are 
less  than  27,  or  3^,  and  more  than  8,  or  2^,  the  tens'  figure  of  the  root  is  2. 
2  tens,  or  20,  cubed  is  8000,  and  8000  subtracted  from  12167  leaves  4167, 
which  is  equal  to  3  times  the  tens2  x  the  units  +  3  times  the  tens  x  the  units2 
+  the  units^. 

Since  3  times  the  tens2  x  the  units  is  much  greater  than  3  times  the  tens 
X  the  units2,  4167  is  a  little  more  than  3  times  the  tens2  x  the  units.  If, 
then,  4167  is  divided  by  3  times  the  tens2,  or  by  1200,  the  trial  divisor,  the 
quotient  3  will  be  the  units  of  the  root,  provided  proper  allowance  has  been 
made  for  the  additions  necessary  to  obtain  the  complete  divisor. 

Since  the  complete  divisor  is  found  by  adding  to  3  times  the  tens2  3  times 
the  tens  x  the  units  and  the  units2,  the  complete  divisor  is  1200  +  180  +  9,  or 
1389.  This  multiplied  by  3  gives  4167,  which  subtracted  from  4167  leaves  no 
remainder.     Therefore,  the  cube  root  of  12167  is  20  +  3,  or  23. 


SECOND    PROCESS 
12-167 

e  =         8 

St""  =1200 

3tu=    180 

w«=       9 

4  167 

1389 

4  167 

23 


Explanation.  —  In  practice  it  is  usual  to 
place  figures  of  the  same  order  in  a  column, 
and  to  disregard  the  ciphers  on  the  right  of  the 
products. 


2.   What  is  the  cube  root  of  1740992427  ? 


«8   = 

3«2  =3(10)2 
3«u  =  3(10  X  2) 
m2  =  22 

= 

1.740.992427 
1 

1203 

0.9 

300 

60 

4 

740 

3  f2  =  3(120)2 

364 
4320 

0 

728 
12992 

12992427 
12992427 

III 

*2. 

3  «2  =  3(1200)2 
3  «M  =  3(1200  X 
m2  =  32 

3)  = 

4320000 

10800 

9 

433080 

9 

Since  a  root  expressed  by  any  number  of  figures  may  be  regarded  as  com- 
posed of  tens  and  units,  the  processes  of  example  1  have  a  general  application. 
Thus,  120  =  12  tens  +  0  units  ;  and  1203  =  120  tens  +  3  units. 


236 


EVOLUTION 


Since  the  third  figure  of  the  root  is  0,  it  is  not  necessary  to  form  the  com- 
plete divisor,  inasmuch  as  tlie  product  to  be  subtracted  will  be  0. 

Rule.  —  Separate  the  numbers  into  periods  of  three  figures  each, 
beginning  at  units. 

Find  the  greatest  cube  in  the  left-hand  period,  and  write  its  root 
for  the  first  term  of  the  required  root.  Cube  the  root,  subtract  the 
result  from  the  left-hand  period,  and  annex  to  the  remainder  the  next 
period  for  a  new  dividend. 

Take  3  times  the  square  of  the  root  already  found,  with  two  ciphers 
annexed,  for  a  trial  divisor,  and  by  it  divide  the  dividend.  The 
quotient,  or  quotient  diminished,  will  be  the  second  figure  of  the  root. 

To  this  trial  divisor  add  3  times  the  product  of  the  first  part  of 
the  root  with  a  cipher  annexed,  multiplied  by  the  second  part,  and 
also  the  square  of  the  second  part.  Tlieir  sum  will  be  the  complete 
divisor. 

Multiply  the  complete  divisor  by  the  second  part  of  the  root,  and 
subtract  the  product  from  the  dividend. 

Continue  thus  until  all  the  figures  of  the  root  have  been  found. 

1.  When  there  is  a  remainder  after  subtracting  the  last  product,  annex 
decimal  ciphers,  and  continue  the  process. 

2.  Decimals  are  pointed  off  from  the  decimal  point  toward  the  right. 

?>.  The  cube  root  of  a  common  fraction  may  be  found  by  extracting  the 
cube  root  of  both  numerator  and  denominator  separately,  or  by  reducing  it 
to  a  decimal  and  then  extracting  its  root. 

Extract  the  cube  root  of 


3. 

29791. 

9.   2406104. 

15. 

.000024389. 

4. 

54872. 

10.   69426531. 

16. 

.001906624. 

5. 

110592. 

11.   28372625. 

17. 

.000912673. 

6. 

300763. 

12.   48.228544. 

18. 

.259694072. 

7. 

681472. 

13.    17173.512. 

19. 

926.859375. 

8. 

941192. 

14.   95.443993. 

20. 

514500.058197. 

Ex 

:tract  the  cube  root  to  three  decimal 

places : 

21 

.   2. 

23.    .8. 

25. 

A- 

27.   f 

22 

.   5. 

24.    .16. 

26. 

2 

28.    ^. 

EVOLUTION  237 

237.   To  extract  any  root  of  a  polynomial. 

To  find  a  formula  for  obtaining  the  complete  divisor  in  extract- 
ing the  fourth,  fijih,  sixth,  or  any  required  root  of  a  polynomial, 
raise  (a  +  b)  to  the  required  power,  and  separate  all  the  terms 
after  the  first  into  two  factors  one  of  which  shall  be  the  second 
term  of  the  root.  The  other  factor  will  be  the  formula  for  the 
complete  divisor. 

(a  +  6)5  =  ffl5  4.  5  a*b  +  10  a^b^  +  10  a^b^  +  bab*  +  b^. 

Trial  divisor,  5  a*. 

Complete  divisor,  5  a*  +  10  a%  +  10  a^b'^  +  5  a¥  +  b*. 

(a  +  6)T  ^  a7  +  7  a<ib  +  21  a^b'^  +  35  a*b^  +  35  a^b*  +  21  a^b^  +  7  ab^  +  b\ 

Trial  divisor,  7  a^. 

Complete  divisor,  7  a«  +  21  a^b  +  35  a^fe^  +  35  a^b^  +  21  a^fc*  +  lab^  +  6«. 

Since  the  fourth  power  is  the  square  of  the  second  power,  the 
sixth  power  the  cube  of  the  second  power,  etc.,  any  root  whose 
index  is  4,  6,  8,  9,  etc.,  may  be  found  by  extracting  successively 
the  roots  corresponding  to  the  factors  of  the  index. 

Tlie  fourth  root  may  be  obtained  by  extracting  the  square  root  of  the 
square  root ;  the  sixth  root,  by  extracting  the  cube  root  of  the  square  root, 
or  the  square  root  of  the  cube  root ;  the  eighth  root,  by  extracting  the  square 
root  of  the  square  root  of  the  square  root. 

Examples 

1.  Find  the  fourth  root  of  16  -  32  a;  +  24  a^  -  8  a^  -f  a^. 

2 .  Find  the  fourth  root  of  a^  + 12  a^y  +  54  a^/  + 108  xf  +  81 2/*. 

3.  Find  the  fourth  root  of  16  m*  -  32  m^  +  24  m^  -  8  m  +  1. 

4.  Find  the  fifth  root  of  32  a;*  +  80  a^  +  80  a^  +  40  x-^  + 10  a;  + 1. 

5.  Find  the  fifth  root  of  a»''+15a8+90a«+270a*+405a2+243. 

6.  Find  the  sixth  root  of  a^  -  12  ar>  +  60  a;^  -  160  a^  +  240  x" 

- 192  a;  +  64. 

7.  Find  the  sixth  root  of  64  a^  -  576  3?  +  2160  x^  -  4320  y? 

+  4860a^2_29i6a;  +  729. 

8.  Find  the  sixth  root  of  a^  +  6  acx'  +  15  a'^c-x''  -|-  20  a^(?x' 

+  15  a''c''Qi?  \  6  aVa;  +  aV. 


238 


EVOLUTION 


238.   To  extract  any  root  of  an  arithmetical  number. 

Examples 
1.   Find  the  cube  root  of  42875. 

Solution 
By  factoring,  42875  =  5x5x5x7x7x7. 


;.  §§  26,  229,  Prin.  3,     Vi2875  =  5  x  7  =  35. 
Find,  by  factoring,  the  roots  indicated ; 

2.  ^/3375. 

3.  a/1296. 


4.  •v/531441. 

5.  ^759375. 


6.  V4084101. 

7.  ^262144. 


8.    Find  the  sixth  root  of  1771561. 
Solution.  — The  square  root  of  1771561  is  1331. 
The  cube  root  of  1331  is  11. 


§  237,  V1771561  =  11. 

Find  the  roots  indicated : 
9.    -^50625.  11.    a/531441. 


10.    V46656. 


12.    V5764801. 


13.  V24137569. 

14.  -v/10604499373. 


239.   Factoring  by  evolution.  , 

Examples 

1.  Factor  a;*  +  4a^  +  8a^  +  8a;-5. 

Solution.  —  Extracting  the  square  root  as  far  as  possible,  the  root  ob- 
tained is  a;2  +  2  a:  4-  2  with  a  remainder  of  —  9. 

Therefore,  a^  +  4a;3  +  8x2  +  8a;-6 

=  (a;2  +  2x4-2)2-9 

§128,  =(x2  +  2a;  +  2  +  3)(x2  +  2x  +  2-3) 

=  (x2  +  2x  +  5)(x2  +  2x-  1). 

Factor  the  following  polynomials : 

2.  iK*  +  6a;«  +  llx2^g3,_8, 

3.  a;«  +  2a;' +6a;^  +  8a;»  +  8ar^  +  8a;  +  3. 

4.  x«-4ar*  f  6a;*  +  6a^-19ic2  +  10a;  +  9. 

5.  4a^  +  12.x'  +  25cc*  +  40a^  +  40ar'  +  32a;  +  15. 


THEORY  OF  EXPONENTS 


240.  Thus  far  the  exponents  used  have  been  positive  integers 
only,  and  consequently  the  laws  relating  to  exponents  have  been 
obtained  in  the  following  restricted  forms : 

1 .  a™  X  a"  =  a""*""  when  m  and  n  are  positive  integers. 

2.  a"*  -r-  a"  =  a™""  when  m  and  n  are  positive  integers  and 
m  >  n. 

3.  (a"*)"  =  a""  when  m  and  n  are  positive  integers. 

4.  -v/o^  =  a"*^"  when  m  and  n  are  positive  integers,  and  m  is  a 
multiple  of  n. 

5.  {ahy  —  a"b"  when  n  is  a  positive  integer. 

If  all  restrictions  are  removed  from  m  and  n,  we  may  then  have 
expressions  like  a~^  and  a^.  But  such  expressions  are,  as  yet, 
unintelligible,  because  —  2  and  f  cannot  show  how  many  times  a 
number  is  used  as  a  factor. 

Since,  however,  these  forms  may  occur  in  algebraic  processes, 
it  is  important  to  discover  meanings  for  them  that  will  allow 
their  use  in  accordance  with  the  laws  already  established,  for 
otherwise  great  complexity  and  confusion  would  arise  in  the  pro- 
cesses involving  them. 

Assuming  that  the  law  of  exponents  for  multiplication, 

a^  X  a''^  «"'+",  (I) 

is  true  for  all  values  of  m  and  n,  the  meanings  of  zero,  negative, 
and  fractional  exponents  may  be  readily  discovered. 

Then,  having  verified  the  remaining  laws  of  exponents  for  these 
exponents,  all  the  laws  will  have  been  shown  to  be  of  general 
application  for  all  commensurable  exponents. 

239 


240  THEORY  OF   EXPONENTS 

241.  The  meaning  of  a  zero  exponent. 

1.  What  is  the  exponent  of  a  in  the  product  of  a^  x  a^?  of 
a^  X  a^  ?  of  a'  x  a**,  if  the  law  of  exponents  for  multiplication  is 
true  for  all  values  of  m  and  n  ? 

2.  Since  a^xa'*  =  a^,  what  is  the  value  of  a**?  What  is  the 
value  of  a^?    of  6«?    of  /?    of  m"? 

3.  What,  then,  is  the  value  of  the  zero  power  of  any  number  ? 

242.  Principle.  —  Any  number  with  a  zero  exponent  is  equal 
to  1. 

The  above  principle  may  be  established  as  follows : 

It  is  to  be  proved  that  a"  =  1. 

Since  Law  I  is  to  be  true  for  all  values  of  m  and  n,  §  240,  when  »  =  0, 

ar^  X  a'^  =  a^+o  =  a™. 

Dividing  by  a",  a"  =  —  =  1. 

a" 

243.  The  meaning  of  a  negative  exponent. 

1.  What  is  the  exponent  of  a  in  the  product  of  a*  x  a^?  of 
a'^  xa?  of  a*  X  a" ?  of  a*  x  a"\  if  the  law  of  exponents  for  multi- 
plication is  true  for  all  values  of  m  and  n?  of  a*  x  a''''  ? 

2.  Since  a*  x  a~^  —  a^  and  since  a*  x  -  =  a%  to  what  fraction  is 

a 

a~^  equal  ?     Since  a*  x  a~^  =  a^,  to  what  fraction  is  a~^  equal  ? 

3.  What  is  another  expression  for  a~^  ?   for  x~^  ?  for  y~*? 

4.  What  is  the  equivalent  of  any  number  with  a  negative 
exponent  ? 

5.  Since  a~^  =  —,  to  what  is  a~^6  equal  ?     ? 

a^  c 

6.  Since  — -  =  1  -^  — ,  or  a^,  to  what  is  - — -  x  - ,  or  — — ,  equal  ? 

a~'^  a^  a  -      c         a~x 

7.  Without  changing  the  value  of  the  fraction,  transfer  a  from 
the  numerator  to  the  denominator  m  ;    in  — ;    from  the 


denominator  to  the  numerator  in      ^ 


h  b 


a~^c         a'c 


THEORY  OF  EXPONENTS  241 

8.  How  may  a  factor  be  transferred  from  one  term  of  a  frac- 
tion to  the  other  without  changing  the  value  of  the  fraction  ? 

244.  Principles.  —  1.  Any  immher  with  a  negative  exponent  is 
equal  to  the  reciprocal  of  the  same  number  with  a  numerically  equal 
positive  exponent. 

2.  Any  factor  may  be  transferred  from  one  term  of  a  fraction  to 
the  other,  without  changing  the  value  of  the  fraction,  if  the  sign  of 
the  exponent  is  changed. 

The  above  principles  may  be  established  as  follows : 

Principle  1.    It  is  to  be  proved  that  a~"  =  — 

a» 

Since  §  240,  Law  I,  a'"  x  a^  =  0™+",  is  to  hold  for  all  values  of  m  and  n, 
when  m  =  —  n,  a-n  x  «»  =  a-"+"  =  a° ; 

but,  §  242,  ao  =  1  ; 

.".  Ax.  1,     /  a"  X  a"  =  1. 

Dividing  by  a",  a  "  =  —  • 

Principle  2.    It  is  to  be  proved  that  ~ —  =  — 

6-»      a™ 

By  Prin.  1,  a-"»  =  —  and  6-»  =  — • 

J_ 

Therefore,  «I"  =  ^  =  ±x^  =  ^. 

b-"     J^     a"      1      a" 

b» 

Examples 
Write  the  following  with  negative  exponents : 

1.  In-a.  3.    1 -h  a".  5.    c~(j^x^. 

2.  1  -T-  a^  4.   a-':-'^?.  Q.   am?  -=-  to". 

7.  Write  ^x~^y^  with  positive  exponents. 
Solution.  —  Prin.  1,        5  x-^j/^  ::3  5  yi     —  £_M_. 

Write  the  following  with  positive  exponents : 

8.  2x-^.  11.   a-'6-\  14.   4aV. 

9.  5a~^  12.    x-^y~^.  15.    3ax~^. 
10.   3  6-".                      13.    a-^b'^c-\  16.    a»6-«". 

ALG. —  iU 


I 


242  THEORY  OF  EXPONENTS 

q   ~2 

17.  Write  — —  without  a  denominator. 

or 

Solution.  —  Prin.  2,  ^  =  3  a'^x-\ 

Write  the  following  without  denominators : 

18.  — .  23.   -X-.-  28.    ^^ 


by  a~^6^  \y 

19.    ^.  24.    A-  29.    r^Y- 


20.  ^.  25.    ^^^'.  30.    (^ 
.  a?h^                                     x^  \bj 

21.  '^.  26.    A.  31.    ^'. 
6^/1^                                   a;"'*  6~' 

22.  — .  ,        27.    — .  32        ^ 


aj"^  ?/  (aby 

245.   The  meaning  of  a  fractional  exponent. 

1.  What  is  the  cube  root  of  a^?  How  is  its  exponent  ob- 
tained ?     Express  the  root  with  this  division  indicated. 

2.  In  a^  what  does  6  express  with  reference  to  a?  What 
does  3  express  with  reference  to  a^  ? 

3.  What  is  the  fifth  root  of  b^^?  Express  the  root  with  a 
fractional  exponent. 

4.  In  b  ^  what  does  15  express  with  reference  to  6  ?  What 
does  5  express  with  reference  to  6'^  ? 

5 .  Express  the  cube  root  of  a  with  a  fractional  exponent ;  the 
fourth  root  of  a^;  the  third  root  of  the  seventh  power  of  a. 

6.  What  does  the  numerator  of  a  fractional  exponent  indi- 
cate ?     What,  the  denominator  ? 

7.  Since  a^  x  a^  X  a^  =  a",  if  the  law  of  exponents  for  multi- 
plication holds  true  for  all  values  of  m  and  n,  of  what  is  a^  the 
cube  root  ? 

Since  a^  x  a^  x  a^  x  a^  x  a^  x  a^  =  a^,  of  what  is  a^  the  sixth 
power  ?     What  two  meanings  may  a*  have  ?    a^?   a^? 

8.  What  does  a  fractional  exponent  indicate  ? 


THEORY  OF  EXPONENTS  243 

246.    Principles.  —  1.    The  numerator  of  a  positive  fractional 
exponent  indicates  a  poiver  and  the  denominator  a  root. 

2.   A  positive  fractional  exponent  indicates  a  root  of  a  power  or  a 
power  of  a  root. 

Prin.  2  refers  to  principal  roots  only  (§  225). 

The  above  principles  may  be  established  as  follows : 

Let  p  and  q  be  any  positive  integers,  and  a  any  positive  number. 

—  p     p    p 
Since  3—  =  -»  -  may  represent  any  positive  fraction. 

p 
1.    It  is  to  be  proved  that  in  a»,  p  indicates  a  power  and  q  a  root. 
Since  Law  I,  or  a"^  x  a"  =  a"»+",  is  to  hold  for  all  values  of  m  and  n, 

when 


also 

?       ?        -  _ 

Hence,        a*  x  a*  x  a*  •••  to  g  factors  =  a «  =  qp. 

Taking  the  gth  root,  §  26  and  Ax.  7, 

p        

a«  =  i/af. 
?  p 

Hence,  in  a',  p  indicates  a  power  and  q  a  root,  and  a'  indicates  the  qth 

root  of  the  pth.  power  of  a. 

E        

2.   It  is  to  be  proved  that         a'  =  v^a",  or  (  \/a)i». 

p         

By  the  previous  proof,  a'  =  VaK  (1) 

I 
It  p  =  l,  a»  =  Va. 

Raising  to  the  pth  power, 
1        1        1 
a'  X  a'  X  a*  •••  to  p  factors  =  (  Vay, 


m 

:  n 

= 

3' 

p 

X 

p 

— 

p  +  p, 
a"    " 

_ , 

2p 

p 

p 

p 

?p 

p 

^ 

«» 

X 

a' 

X 

a» 

= 

a« 

X 

a« 

= 

ffl«, 

etc. 

or,  Law  I,  o«=(Va)J'.  (2) 

p 

Hence,  (1)  and  (2),  a'  indicates  the  gth  rooi  0/  «/ie  pth  power  of  a,  or  the 

pth  power  of  the  qth  root  of  a. 

-?      1 
It  follows  from  §  244  that  a   »  =  — • 

p 

a' 


244  THEORY  OF  EXPONENTS 


Examples 


1.  Express  s/a'bc'*  with  positive  fractional  exponents. 
Solution.  \/a^bc^*  =  ahh~^  =  ^^^. 

Express  with  positive  fractional  exponents : 

2.  Va6^.  5.    (V^)^  8.  (^/a^)-2. 

3.  -Vxy.  6.    {-^yy.  9.  5Vx-hj-\ 

4.  V^.  7.    (^a&)3.  10.  2-v/(a  +  6)=^. 

In  the  following,  large  numbers  may  be  avoided  by  extracting  the  root  first. 
Find  the  value  of 

11.  8^.  15.    64^  19.    64"l 

12.  si  16.    32l  20.    (-8)"^. 

13.  8"*.  17.    25\  21.    (-32)-i 

14.  (-8)1  18.   8ll  22.    16"*. 

Simplify : 

23.  ^/c^'  +  x^  +  8^+3x^-5^x-'^/2r. 

24.  4.^x  +  5af-Sx-^  +  2\/^'-8^-2xi 

25.  -v/^--v/a^+-^a6^_5_^^_^4^^_4^i5§^^_ 

Express  with  radical  signs : 

26.  ak  29.    a^fti  32.  a^ -^  .t^ 

27.  x^.  30.    iK^i  33.  m^-T-nk 

28.  .^'i  31.    aV^.  34.  .^•^ -=- ?/! 

247.  It  now  remains  to  complete  the  proof  that  the  other  laws 
of  exponents  are  of  general  application  for  commensurable  ex- 
ponents by  showing  that  they  apply  when  negative^nd  fractional 
exponents  are  employed  with  the  meanings  just  obtained. 


THEORY  OF  EXPONENTS  245 

248.  To  prove  that  the  law  of  exponents  for  division  is  general. 

It  is  to  be  proved  that  a"^  ^  a'^  =  a™-»  for  all  values  of  m  and  n. 

Since  dividing  by  a"  is  equivalent  to  multiplying  by  its  reciprocal  — , 

a" 

§  106,  o™  -r-  a"  =  «"  X  — 

§  244,  Prin.  2,  =  a"  x  a"". 

Hence,  for  all  values  of  m  and  ?t,   a'"  ^  a^  =  a"*"".  (II) 

249.  To  prove  that  the  law  of  exponents  for  involution  is  general. 

It  is  to  be  proved  that  (a™)"  =  a™"  for  all  values  of  m  and  n. 

Case  1 .  —  Let  m  represent  any  value  and  n  a  positive  integer. 

Then,  (a"*)"  =  «"•  x  a"*  x  a"*  •••  to  w  factors- 

Law  I,  =:  Qm+m+m  ...  to  n  terms 

P 

Case  2.  —  Let  m   represent  any  value,  and  let  ?j  =  — ,  p  and  q  being 
positive  integers. 


Then,  §  246,  Prin.  2,  (a")?  =  </{a'^)p 

Case  1,  =  Va^ 

mp 

§  246,  Prin.  1,  =  aJ. 

Case  3.  —  Let  m  represent  any  value,  and  let  n  =  —  r,  r  being  a  positive 
integer  or  a  positive  fraction. 

1 


Then,  (a™)" 


(«")' 


Case  1,  =— _ 

a""- 

§  244,  Prin.  2,  =  a-""". 

Hence,  for  all  values  of  m  and  n,  («"")"  =  a"*".  (HI) 

250.    To  prove  that  the  law  of  exponents  for  evolution  is  general. 

It  is  to  be  proved  that   y/a""  =  a"*^"  for  all  values  of  m  and  n. 

Since  (a™)"  =  a""'  for  all  values  of  m  and  n,  it  is  true  when    ~  is  sub- 

71 

stituted  for  n. 

1  -  mx- 

Substituting  -  for  n,  (a*")"  =  a"  ", 

or,  §  246,  Va""  =  a™^".  (IV) 


246  THEORY  OF  EXPONENTS 

251.    To  prove  (ab)"  =  a"b"  for  all  values  of  n. 

P 
Case  1.  — Let  n  =  -,  p  and  q  being  positive  integers. 

Then,  §  249,  Case  1,  since  g  is  a  positive  integer, 

[(a6)»]«=(a&)«''*=(a6)p 
§  240,  5,  =  aPhP.  (1) 

Also,  §  249,  Case  1, 

p  p  p  p       p  p 

{a^h^Y  =  a»6»  x  a«6«  •••  to  q  factors 

p       p  p       p 

§  83,  =  (a«  X  a«  •••  to  g-  factors) (6*  x  6«  •••  to  5-  factors) 

=  aPhP.  (2) 

(1)  and  (2),  Ax.  1,  [(a&)']»  =  (a»6»)«. 

p       p  p 
Taking  the  qih  root,  (a6)«  =  a^6«. 

Case  2.  — Let  «  =  —  r,  »•  being  a  positive  integer  or  a  positive  fraction. 
Then,  §  244,  Prin.  1,  (aft)"''  =  — ^ 

Case  1,  =  -i- 

§  244,  Prin.  2,  =  a-'-fi-'. 

Hence,  for  all  values  of  n,   (a&)"  =  a"6".  (V) 


Examples 
252.   Multiply: 

1.  a^  by  a-2.  3.   a*  by  a"*.  5.   a^  by  a". 

2.  a^  by  a-^  4.   a  by  a~*.  6.   a;^  by  x^. 

7.  a^6^  by  a^si  •  10.   w"^  by  ani 

8.  m^n  by  m^w~\  11.    a"""  by  a""''. 

OT+n  w»-« 

9.  a*&'  by  a~*6*.  12.    a  ^    by  a  ^  . 

13.  Multiply  x^y"^  +  cc*  +  x^y^  -f  a;^?/   +  2/    by  a;*?/*. 

14.  Multiply  2/"  +  x-y +' +  a;- V+2  +  a;-32/"+3  by  ary-\ 


THEORY  OF  EXPONENTS  247 

15.   Expand  (ah~^ -^1  +  arh^){ah~^ -1  +  a'h^. 

First  Solution 

a^b~^  +      1      +  a~h^ 
ah~^  -      1      +  a~h^ 

ah-}  +  ah~^  +  oOfeo 

-ah~^-    1    -a~h^ 

+  a^b'^  +  a~^b^  +  a^b 

ah-^  +1  +  a~^b 

Second  Solution 

(ah~^  +  1  +  a~h^)(ah~^  -  1  +  a'h^) 

=  ah-^  +  2  a^fto  +  a'h  -  1 
=  ah-^  +  2  +  a'h  -  1 
=  ah-^  +  1  +  a~h. 
Expand : 

16.  {a^+b^)(a^  —  b^).         21.    {x^—xhj~^+y~^){x^  +  y~^). 

17.  {x^+y^)(x^—y^).         22.    {a^  —  b~^+ah~^  +  l)(a^-b^). 

18.  (x-^  +  10){x-^-l).       23.    (l-x+o^)(x-^+x-^+x-'). 

19.  (a;^-4)(a:^+5).  24.    (a-i4-&~^+c^)(a-^  +  &~^+2c*). 

20.  (x*-?/*)(x*+2/^).         25.   {a'b'-ab^+b^a-'b-^+a-^b-'+a-*). 

Divide : 

26.  a'  by  a*.  28.   a'^  by  a-\  30.    a;^  by  a;i 

27.  a^  by  a".  29.    x^  by  ic~i  31.    a;""^  by  a;"-". 

32.  Divide  a:*  +  a;^?/^  +  i/*  by  a^^t/^. 

33.  Divide  a  "»  +  a-^'d  +  fe''  by  a-^b. 

34.  Divide  a^  +  2  a»^  +  3  a^ar^  +  a^a;  —  a*  by  a*x*. 


248  THEORY  OF  EXPONENTS 

y/    35.    Divide  b'' +  S  a"'^  -  10  a'^b  by  ah-^-2. 

Solution 

aVi  -  2)6-1  +  3  a"^  -  10  a-^b(a~^  +  5  a-^b 
6-1  -  2  a~^ 


5  a  2-10  «-i6 
5  a~2  _  10  a-16 


Divide : 

36.  a-b  by  a^  +  bK  41.    J -{- 2  ah~^  +  b'^  hy  a^  +  b~k 

37.  a  —  6  by  a^  —  b^.  42.    x^  —  2  +  x~^  by  a;^  —  a;"^. 

38.  a +  6  by  a^  +  fti  43.    3-4a;-^  +  a;-2  by  o;-^  —  3. 

39.  a^  +  b^  by  a^  +  b^.  44.    4  x^i/-^  —  5  ?/  +  a?  V  by  iK^  —  2/^. 

40.  x—1  by  a;^  +  a;^  +  1.  45.    a^—  W  by  a^  +  b^. 

Simplify  the  following : 

46.  {a^y.  49.  (-a^)».  52.    (8-*)*. 

47.  (a~^)^  50.  {-a?)\  53.    (16"^)^. 

48.  (a-'')2.  51.  {-a^)-\  54.    (- ^V)"^. 

Expand  by  the  binomial  formula : 

55.  {a^-b^y.  57.  (a-'- b^y.  59.    (a~*  +  i)^ 

56.  (a*  +  6^^  58.  (a;"2-y^y.  60.    {1  —  x^y. 

Simplify  the  following : 


61.  \xi.  64.    Va~^6  ^  67.    {■^^a'h^)'^. 

62.  Va"*.  65.    V^.T%-3.  68.    (^m'hr^)k 

63.  Va"^.  66.    ^aFb".  69.    (4aj2»?/-V)i 
Extract  the  square  root  of 

70.  a.-^  +  2  a;^  +  3  a;  +  4  a;^  +  3  +  2  a;~2  4-  x-\ 

71.  ar^  +  ?/  +  4  2;~2— 2 a;?/^  +  4  .tz"'  —  4 y^z-^ 

72.  a  +  4  &*  +  9  c^  -  4  a^^^  +  6  ah^  -  12  ^^c*. 


THEORY  OF  EXPONENTS  249 

Extract  the  cube  root  of 

73.  a2  +  6a^-f  12a*  +  8. 

74.  a-^ah^  +  ^ah^  -h\ 

75.  ^x-^  —  \2x~iy+&x'^y--f. 

76.  x^~&x  +  in  x^  -20  +  15  x'^  -  6  a;-i  +  x~^. 

77.  Factor  4a;~^  — 9y~^,  and  express  the  result  with  positive 
exponents. 

Solution 

§  128,  4  x-2  -  9  y-2  =  (2  a;-i  +  3  y-^)  (2  x"!  -  3  y-i) 

_/2      3\/2_3\ 

~U    2//U    y) 

Factor  the  following  and  express  the  results  with   positive 
exponents : 

78.  a-^  —  h-\  83.  a^  -  a;"*. 

79.  9-x-\  84.  o?  +  2  +  a\ 

80.  16 -a-*,  j  85.  h'-S  +  lQh-\ 

81.  27-6-3.  86.  12-x-^-x-\ 

82.  h-'^  +  y-^  87.  2  -  3  a;"' -  2  a;-3. 

Solve  the  following  equations : 

88.    x^  =  2.  96.  a."^  =  6. 

97.  a;~^  =  16. 

98.  25  a;-*  =  1. 

99.  x2=:243. 

100.  a;^  +  32  =  0. 

101.  x^  +  a^  =  0. 
94.  ^a;^  =  25.  102.  a;^  -  64  =  0. 
96.    2a;"^  =  ^.                                   103.  a;~^  +  27  =  0. 


89. 

3^-^  =  8. 

90. 

x^  —  4. 

91. 

x^  =  16. 

92. 

ix^  =  9. 

93. 

a;-^  =  5. 

RADICALS 


253.  1.   What  is  indicated  by  V«?  by  a;^^  by -^9  by  ah 

2.  Indicate  in  two  ways  the  square  root  of  25 ;  of  36 ;  of  2 ; 
of  3. 

3.  Which  of  these  indicated  roots  can  be  obtained  exactly  ? 
Which  cannot  be  obtained  exactly  ? 

254.  An  indicated  root  of  an  expression  is  called  a  Radical. 
The  root  may  be  indicated  by  a  radical  sign  or  by  a  fractional 

exponent. 

\/5^,  (5a)i,  \/a^,  («a;6)3,  Va^  +  2ab  +  b^,  and  {a"^  +  2  ab  +  b^)^  are 
radicals. 

In  the  discussion  and  treatment  of  radicals  only  principal  roots 
will  be  considered. 

255.  The  Degree  of  a  radical  is  indicated  by  the  index  of  the 
root  or  by  the  denominator  of  the  fractional  exponent. 

Va  +  X  and  {b  +  xy  are  radicals  of  the  second  degree. 

256.  When  the  indicated  root  of  a  rational  number  cannot  be 
exactly  obtained,  the  expression  is  called  a  Surd. 

V2  is  a  Surd,  since  2  is  rational  but  has  no  rational  square  root. 
V  1  +  V3  is  not  a  surd,  because  1  +  VS  is  not  rational. 

Radicals  may  be  either  rational  or  irrational,  but  surds  are 
always  irrational. 

Both  Vi  and  VS  are  radicals  but  only  V3  is  a  surd. 

257.  An  indicated  even  root  of  a  negative  number  is  called  an 
Imaginary  Number ;  as  V—  4,  V  —  a. 

All  other  numbers,  whether  rational  or  irrational,  are  called 

Heal  Numbers;  as  V25,  VS,  a^,  4. 

250 


RADICALS  251 

258.  A  surd  may  contain  a  rational  factor,  that  is,  a  factor 
which  is  a  perfect  power  of  the  same  degree  as  the  radical.  The 
rational  factor  may  be  removed  and  written  as  the  coefficient  of  the 
irrational  factor. 

In  VS  =  V4  X  2  and  \/54  =  \/'21  x  2,  the  rational  factors  are  V4  and 
\^;  that  is,  VS  =  2V2  and  v^54  =  3v^. 

259.  A  surd  that  has  a  rational  coefficient  is  called  a  Mixed 
Surd. 

2V2,  aVic^,  and  (a  —  b)  Va  +  b  are  mixed  surds. 

260.  A  surd  that  has  no  rational  coefficient  except  unity  is 
called  an  Entire  Surd. 

VS,  VU,  and  \/cfi  +  a;^  are  entire  surds. 

261.  A  radical  is  in  its  simplest  form  when  the  expression 
under  the  radical  sign  is  integral,  contains  no  factor  that  is 
a  power  of  the  same  degree  as  the  radical,  and  is  not  itself  a 
perfect  power  whose  exponent  is  a  factor  of  the  index  of  the 
radical. 

V?  is  in  its  simplest  form ;  but  Vf  is  not  in  its  simplest  form,  because 
I  is  not  integral  in  form  ;  VS  is  not  in  its  simplest  form,  because  the  square 
root  of  4,  a  factor  of  8,  may  be  extracted  ;  \/25,  or  25^,  is  not  in  its  simplest 
form,  because  25^  =(52)^  =  5^  =  5^,  or  \/5. 

REDUCTION  OF  RADICALS 

262.  To  reduce  a  radical  to  its  simplest  form  when  it  has  a  rational 

factor. 

Examples 


1.    Reduce  V20a^  to  its  simplest  form. 

process 


V20a«  =  V4a«  x5  =  V4^«  X  VS  =  2a^V5 

Explanation.  —  Since  the  highest  factor  of  20  (fi  that  is  a  perfect  square 
is  4a^,  V^Oo^  is  separated  into  two  factors,  a  rational  factor  ^4 a**,  and  an 
irrational  factor  V5.  y/20a^  =\/4a*'  x  V5,  §  229,  Prin.  3.  Extracting  the 
square  root  of  4  a^  and  prefixing.the  root  to  the  irrational  factor  as  a  coeflBcient, 
the  result  is  2  a^VS. 


252  RADICALS 

2.   Keduce  V^-  864  to  its  simplest  form. 

PROCESS 

^37864=^/- 216  x  4  =\/^^216  x  ^/4  =  - 6^ 

RuLK.  ^-  Separate  the  radical  into  two  factors  one  of  which  is 
its  highest  rational  factor.  Extract  the  required  root  of  the  rational 
factor,  multiply  the  result  by  the  coefficient,  if  any,  of  the  given 
radical,  and  place  the  product  as  the  coefficient  of  the  irrational 
factor. 

Simplify  the  following : 


3. 

V12. 

4. 

V75. 

5. 

^16. 

6. 

V128. 

7. 

^250. 

8. 

^32. 

9. 

V600. 

10. 

V500. 

11. 

a/160. 

12. 

^3000. 

13. 

^81. 

14. 

v'189. 

15.    V162.  27.    V243a^ri' 


16.  Vl8a2.  28.    V128a%^ 

17.  V256.  29.    V405 aY. 


18.  ^Wc\  30.  V375a;Y. 

19.  V50a.  31.  (245 a V')^- 

20.  a/640.  32.  (135a;V)l 

21.  V84.  33.  (a^  +  Sa^)*. 


22.  a/72.  34.    (16x-16)^. 

23.  a/192.  35.    Vl8  re  -  9. 

24.  V800.  36.    ^W-^^2^. 


25.  V3645.  37.    a/8-20  61 

26.  a/735.  38.    5  V4  a-  +  4. 

39.  -\/bx^-10xy  +  b  y-.  41.    (3  am- +  6  am  +  3a)i 

40.  a/4  a^  -  24  a^o;  +  36  ax\     42.    (x*?/ -  3  aj^^^  +  3  a^^/^  _  a.y4)i 


/  a 

43.    Reduce  a/ to  its  simplest  form. 

\2f 


PROCESS 


^/Z  =^r«^5IZ  ^  Jj^  X  V2^  = -^  a/2^ 


RADICALS  253 

Explanation.  — Since  a  radical  is  not  in  its  simplest  form  when  the  ex- 
pression under  the  radical  sign  is  fractional,  the  denominator  is  to  be  removed ; 
and  since  the  radical  is  of  the  second  degree,  the  denominator  must  be  made 
a  perfect  square.  The  smallest  factor  that  will  accomplish  this  is  2  y.  Multi- 
plying the  terms  of  the  fraction  by  this  factor,  the  largest  rational  factor 

of  the  resulting  radical  is  found  to  be  \/-^,  which  is  equal  to  -^-     There- 

\  4  y4  2  y'^ 

fore,  the  irrational  factor  is  v'2  y,  and  its  coefficient  is  — -• 

'  "'  2y^  ' 

Simplify  the  following : 

"■    ^i-  '  62.    yf^.  66     -  '  2 


45. 

vi- 

46. 

vi. 

47. 

Vi. 

48. 

V|. 

49. 

V|. 

50. 

^A- 

51. 

n 

3f 

54.    ^'£  58. 

Mil 


3x 


60. 


50  a^y 


61        2.V        W^^  3     (g  +  hf  sj  a  +  b 

x-2y^     2y  '      a-b    \l (a  -  bf 

263.  To  reduce  a  radical  to  its  simplest  form  when  the  expression 
under  the  radical  sign  is  a  perfect  power  of  a  degree  corresponding  to 
eome  factor  of  the  index  of  the  root. 

Examples 

1.    Reduce  v9a^  to  its  simplest  form, 

PROCESS 

-^/9^^  =  V(3^y  =  (3  a)^  =  (3  a)^  =  ^3^ 
5$.    Keduce  V64a*'6'^  to  its  simplest  form. 

PROCESS 

^/6iaFb^'  =  ^/^^^W*  =  6  (2  ab)'^  =  b(2  ab)^  =  b  ^la^ 


254  RADICALS 

Simplify  the  following : 

3.  a/36.  7.  a/1600.  11.  ^'9  d'U'c\ 

4.  a/25.  8.  A/27a8.  12.  a/121  aV. 

5.  A/ii4.  9.  A^343.  13.  Va%^c*d\ 

6.  a/81.  10.  ^289.  14.  -^{x'-2xy-\-y^. 

264.   To  reduce  a  mixed  surd  to  an  entire  surd. 

Examples 
1.   Express  2  a  a/5  h  as  an  entire  surd. 

PROCESS 


2a  a/66  =  V4^  a/56  =A/4a2x5&=A/20^ 

Rule.  —  Raise  the  coefficient  to  a  power  corresponding  to  the 
index  of  the  given  radical,  and  introduce  tJte  result  under  the  radical 
sign  as  a  factor. 

Express  as  entire  surds : 

2.   2  a/2.  6.   3  a/3.  10.   ^V2.  14.   i^/^. 


3.  3V5.  7.    4  a/5.  11.    fV^.  15.    fA/ffa^. 

4.  5a^.  8     I  a/8.  12.    ^Vbc.  16.    I^H. 

5.  3  a/2.  9     a2A/6.  13.    fV|.  17.    |^. 


18.   ^±^^/^H5.        19.    ^^±iJl §_.        20.    l(a-6)l 

265.  To  reduce  radicals  of  different  degrees  to  equivalent  radicals 
of  the  same  degree. 

1.  Express  a^  by  an  equivalent  radical  with  an  exponent  in 
higher  terms. 

2.  What  is  the  degree  of  the  radical  x^?  Express  x*  as  a 
radical  of  the  12th  degree.  Express  x^  as  a  radical  of  the  12th 
degree.     Express  6^  and  6*  as  radicals  of  the  same  degree. 

Examples 

1.  Reduce  a^3,  a/2,  and  a/4  to  equivalent  radicals  of  the 
same  degree. 


RADICALS  255 

PROCESS 

</3  =  3^  =  3A=^3^  =  ^27 
V2  =  2i=2A  =  ^2"«=^^ 
^  =  4*  =  4^^  =  ^^4*  =  ^256 

Rule.  —  Express  the  given  radicals  with  fractional  exponents 
having  a  common  denominator. 

Raise  each  number  to  the  power  indicated  by  the  numerator  of  its 
fractional  exponent,  and  indicate  the  root  expressed  by  the  common 

denominator. 

I 

Reduce  to  equivalent  radicals  of  the  same  degree : 

2.  V2  and  ^/3.  9.  Va&,  -^ab",  and  ^/2. 

3.  V5  and  ^s/G.  10.  Va,  V&,  Vx,  and  Vy. 

4.  -y/T  and  VIO.  11.  ^/a  +  b  and  Vic  +  t/. 

5.  -v^,  V2,  and  ^5.  12.  V|,  ^v^S^,  and  2V5. 

6.  V4,  V2,  and  V3.  13.  -\/x,  y/xy,  and  ■y/x^y'^. 

7.  "\/l3,  VS,  and  V4.  14.  (a  +  &)Va— &  and  y/a  —  b. 

8.  V3,  ^5,  and  ^/J^.  15.  Va+&,  \/a2+62,  and  Va^. 

ADDITION  AND  SUBTRACTION  OF  RADICALS 

266.  Radical  terms  that,  in  their  simplest  forms,  are  of  the 
>ame  degree  and  have  the  same  number  under  the  radical  sign 
are  called  Similar  Radicals. 

267.  Principle.  —  Only  similar  radicals  can  be  united  into  one 
term  by  addition  or  subtraction. 

Examples 

1.   Find  the  sum  of  V50,  2^/8,  and  6VJ. 

Explanation.  —  To  ascertain  whether  or  not  the  given 
-v/50  =    5  V2     expressions  are  similar  radicals,  each  may  be  reduced  to 

o  6/q 9/9     **^  simplest  form.     Since,  in  their  simplest  form,  they  are 

_  of  the  same  degree  and  have  the  same  number  under  the 
6  V ^  =  3  V 2  radical  sign,  they  are  similar,  and  their  sum  is  the  sum 
Sum  —  1 0  a/2     ^^  ^^^  coeflScients  prefixed  to  the  common  radical  factor. 


266  RADICALS 

Find  the  sum  of 
'      2.    ^^50,  Vis,  and  V98. 

3.  V27,  Vl2,  and  V75. 

4.  V20,  V80,  and  ViS. 

5.  V28,  V63,  and  VTOO. 

10.  V|,   Vi2l,  Vl,  and  ViJ. 

11.  V|,  VT5,  |V3,  and  VT2. 

12.  ^%  iV3,  1^9,  and  Vl47. 

13.  ViO,  V28,  -v^25,  and  VT75. 

14.  a/375,  a/44,  V192,  and  a/99 

Simplify  : 

15.  a/245- a/405 +V45. 

16.  V 12  +  3  V75  -  2  v/27. 

17.  5V72  +  3a/18-a/50. 

18.  Vr28  +  v/686-A/54. 

19.  A/IT2-V341+ a/448. 

20.  Vl35- a/625  4- a/320. 

21.  vi+^i+<m-  _ 

22.    V864- a/4000  +  a/32 


6.  a/250,  a/T6,  and  A^Si. 

7.  A/i28,  a/686,  and  Vi- 

8.  a/135,  a/320,  and  ^625. 
9. 


500,  a/108,  and  V- 32. 


23.    A/l28a;+V375a;-A/54x. 


y 


25. 


V^i-v 


16  aar^  ,     /4  aa^ 


A/(a  +  &)2c-A/(a-&)«c, 


27 


28.  6Vff  +  4^|f-8VfH- 

29.  </'-  96  a-^  +  2 a/31?  -  a/5^-  +  a/40^. 

30.  Vabx  -  V«^&^  +  a/8  a^'fis^^. 


31.    A/3a?  +  30«2^75a._V3a^-6a^  +  3a;. 


32.    a/5  a^  +  30  a*  +  45  a^  -^baP-  40  a*  +  80  a\ 


33.  A/50+V9-4Vi+A/^=24+A/27-V64. 

34.  a/|  +  6V|- ^18+ a/36- A/if  + ^125-2 VS- 


RADICALS  257 

MULTIPLICATION   OP   RADICALS 

268.  1.  What  is  the  exponent  of  a  in  d^  x  a^?  in  a^  x  a^? 
in  a^  X  a^?   in  a*  x  a*  ?   in  a^  x  a»  ?  in  Va  x  -v^a  ? 

2.  When  the  fractional  exponents  indicate  different  roots,  what 
must  be  done  before  the  radicals  can  be  multiplied  together  ? 

Examples 
PEOCESSEs.  —  1.    V7xV5=V35 

2.  5V3x2Vi5  =  10V45  =  10x3V5  =  30V5 

3.  2 V3  X  3^  =  2a/27  X  3-^^  =  6\/i08 

Rule.  —  If  necessa'ry,  reduce  the  radicals  to  the  same  degree. 

Multiply  the  coefficients  together  for  the  coefficient  of  the  product 
and  the  factors  under  the  radical  sign  for  the  radical  factor  of  the 
product,  and  simplify  tlie  result. 

Multiply : 

4.  V2  by  Vs.  '  13.  2V6  by  Vl8. 

5.  V2  by  V6.  14.  2^3  by  3^46. 

6.  V3  by  Vis.  15.  2V6  by  3V6. 

7.  V3  by  V48.  16.  3V3  by  2^6. 

8.  2V5  by  3VI0.  17.  </5  by  ViO. 

9.  3V20  by  2V2.  18.  2^250  by  V2. 

10.  V2  by  3V3.  19.   2V24  by  VlS. 

11.  V2  by  2V5.  20.    V28  by  3V7. 

12.  V3  by  3V3.  21.   2V2  by  V512. 

Find  the  value  of 

22.  Va6  X  Vbc  x  Vcd  x  Vda. 

23.  VaV  X  Vl2^  X  V75^. 

24.  V2^  X  Voftc  X  ^/4a^. 

ALG.  17 


258  RADICALS 

85.  Vm7i  X  ^Ttihx  X  ^mn*. 

26.  V2  axy  x  Va;?/  x  -\/a?xy. 

27.  ^x~^y  X  Va;~V  x  -Vx~^y^. 

28.  Va^=^  X  \/a262  x  </(a  -  6)-2. 

29.  V|xVfxV|.  32.   ^rx^fxVf. 

30.  VixV|xV|.  33.    ^jx^xV|. 

31.  ^1  X  ^f  X  VJ.  34.  .  \/|  X  V|  X  ^|. 
35.    Multiply  2V2  +  3V3  by  5V2-2V3. 

Solution 

2V2  +  3V3 
5V'2-2>/3 

20  +  15V6 
-    4\/6-18 

20+  iiVe-is 

=    2+ll\/6. 

36.  V5+V3  by  V5-V3. 

37.  V7  +  V2  by  V7  -  V2. 

38.  V6-V5  by  V6-V5. 

39.  5-V5  by  1  +V5. 

40.  4V7  +  1  by  4V7-1. 

41.  2V2+V3  by  4V2+V3. 

42.  2V3  4-3V5  by  3V3  +  2V6. 

43.  Sa+VS  by  2a  — V5. 

44.  2V6-3V5  by  4V3-ViO. 

45.  a"-  a6V2  +  b^  by  a"  +  a6V2  +  b\ 

46.  a:  —  Vx^  +  yz  by  Va;  +  V.y2. 

47.  xy/x  —  a; Vi/  +  yVx  —  2/V2/  by  Va;  +  Vy. 


Multiply ; 


RADICALS  259 

DIVISION  OP  RADICALS 

269.  1.  What  is  the  exponent  of  a  in  a^-j-a^?  in  a^-^a^? 
in  a^-i-a^?   in  cC^-i-a^?   in  ^a^-^a2 

2.  When  the  fractional  exponents  indicate  different  roots,  what 
must  be  done  before  one  radical  can  be  divided  by  another  ? 

Examples 
PROCESSES.  —  1 .    V60  -7-  Vl2  =  V5 

2.    ^/2H-V2=^4-^8=^|  =  ^||=^^/32 

Rule.  —  If  necessary,  reduce  the  radicals  to  the  same  degree. 

To  the  quotient  of  the  coefficients  annex  the  quotient  of  the  radical 
factors  under  the  common  radical  sign,  and  reduce  the  result  to  its 
simplest  form. 


Find  the  quotient  of 

4.    V50-hV8. 

12. 

2v^-r-V8. 

5.    V72--2V6. 

13. 

-\/ax  -^y/xy. 

6.   4V5-J-V40. 

14. 

^/2ab^^</a*b\ 

7.   6V7H-V126. 

15. 

-\/a^3^  -j-  V2  ax. 

8.    ^4--V2. 

16. 

^9a'b^-^^3ab. 

9.   7^/135-^/9. 

17. 

\/4  a^y^  -7-  \/2  xy. 

10.   7V75H-5V28. 

18. 

Va  —  6  -T-  Va  +  6. 

11.    ■v/l6^v/32. 

19. 

s^^Vf. 

20.  Divide  VIS-VS  by  V3. 

21.  Divide  a/6-2V3  +  4  by  V2. 

22.  Divide  V2  +  2  +  ^  V42  by  ^  V6. 

23.  Divide  5V2  +  5V3  by  Vl04-Vi5. 

24.  Divide  5  +  5V30  +  36  by  V5  +  2V6. 


260 


RADICALS 


INVOLUTION  AND  EVOLUTION  OP  RADICALS 

270.    In  finding  powers  and  roots  of  radicals  it  is  frequently 
convenient  to  use  fractional  exponents. 

Examples 

1.  What  is  the  cube  of  2V'ttx^? 

Solution.      (2Vax^Y  =  2^(ax^)^  =  8  a^x^  =  SVcM  =  8  aX^Vax. 

2.  What  is  the  square  of  3  V^  ? 

Solution.      (3v^)2  =  9(a;5)t  =  9x^  =  9v^  =  Qxv^x^. 

3.  What  is  the  cube  of  V2  +  1  ? 

Solution 
(\/2  +  l)8=(V^)3  +  8(V2)2.'l  +3V'2.12  +  1» 
=  2V2  +  6  +  3V2  +  1 
=  7  +  5V2. 
In  such  cases  expand  by  the  binomial  formula. 


Square : 

4.  3Va6. 

5.  2^^^. 

6.  x^/2^. 

7.  nWlb. 

8.  a-y/a^b. 

Expand : 

19.  (2+V6)== 

20.  (2-|-V2)2 


Cube: 

9.   2V5. 

10.  3V2. 

11.  2^/^. 

12.  \/a^. 

13.  ^4^«. 


Involve  as  indicated; 

14.  (-2V2a6)*. 

15.  (-■y/2y/xy. 

16.  (-V2\/'a^)*. 

17.  (-2V^^)«. 

18.  (-3aV)«. 


22.  (2-Vsy.  25.    (Vx±iy. 

23.  (V7-^/6)l  26.    (Va-Vbf 
21.    (2+V5)3         24.    (2V2-V3)2          27.     Vx±lf. 
28.    What  is  the  cube  root  of  —  27 Vox? 

Solution.  -^ -27Vax.=  i- 27)\ax)^  =-SVax. 


RADICALS  261 

29.  What  is  the  fourth  root  of  -y/^x ? 
Solution.  ^V2x  =[(2x)^]i  =(2a;)^  =\/2«. 

Find  the  square  root  of  Find  the  cube  root  of 

30.  V2.         33.    -y/^.  36.    ^2x.  39.    -  27 V^. 

31.  ^/5.         34.    \/x^\  37.    V7V.  40.    -64a/^ 

32.  ^'^.         35.    VaV.  38.     v/SmV.       41.    —  Vo^. 

Simplify  the  following  indicated  roots : 


42.  V-v/4aV.  44.    (V8aV)i 

43.  'VVa^a^.  45.    (vVy»)"*". 


46. 


RATIONALIZATION 


1  ^72 

271.  How  may  the  fraction  be  reduced  to  the  form  -^— ? 

•^  _       V2  2 

A  to  ^^    -^  to  2V5^    -^  to  ^? 
V2  2  V5  5  vx  a; 

272.  The  process  of  rendering  a  surd  expression  rational  is 
called  Rationalization. 

The  factor  by  which  a  surd  expression  is  multiplied  to  render 
it  rational  is  called  the  Rationalizing  Factor.  * 

The  denominator  of  — -  is  rendered  rational  by  multiplying  it  by  Vs. 

^         V3  ♦ 
2        2v3        /— 
Thus,  —  = .     v3  is  a  rationalizing  factor  for  the  denominator. 

\/3        3 

273.  A  binomial,  one  or  both  of  whose  terms  are  surds,  is  called 
a  Binomial  Surd. 

2  +  V3,  Vx  +  Vy,  and  a*  —  6'  are  binomial  surds. 

274.  Two  binomial  surds  of  the  second  degree  whose  product 
is  rational  are  called  Conjugate  Surds. 

Vo  +  Vb  and  Va  —  Vb  are  conjugate  surds ;  also  a  +  y/b  and  a  —  Vb. 
Conjugate  surds  differ  only  in  the  sign  of  one  of  the  terms. 


262  RADICALS 

275.  Principle.  — A  binomial  surd  of  the  second  degree  may  he 
rationalized  by  miUtiplying  it  by  its  conjugate. 

Thus,  the  product  of  a  +  Vfc  and  a  —Vb  is  a^  —  b. 

Examples 

1.  Find  the  simplest  rationalizing  factor  for  V3a;. 
Solution.  -k/Sx  x  VS x  =  ^9 a;"^  =  3 x. 

.•.  y/Wx  is  the  simplest  rationalizing  factor. 

2.  Find  the  simplest  rationalizing  factor  for  ■v^4  a. 
Solution.  \/4  o  x  v^2  a^  =  y/8  a^  =  2  a. 

,-.  V2  a^  is  the  simplest  raiionalizing  factor. 

Find  the  simplest  rationalizing  factor  for 

3.  V3.  5.    Vox,  7.    V9^.  9,    \/4. 

4.  V6.  6.    vTa.  8.    Vs.  10.    ^/oF. 

11.  Find  the  simplest  rationalizing  factor  for  Vo  +  V3. 
Solution.     (  V5  +  \/3)(V5  -•\/3)  =  (\/5)2_(  V3)2  =  5  -  3  =  2, 

.•.  \/5  —  y/3  is  the  simplest  rationalizing  factor. 

Explanation.  —  Since  (^5)2  =  5  and  (\/3)2  =  3,  Vd  +  VS  may  be 
rationalized  by  multiplying  it  by  some  factor  that  will  give  the  square  of 
each  term,  but  no  other  terms.  Since  the  product  of  the  sum  and  difference 
of  two  numbers  is  equal  to  the  difference  of  their  squares,  the  simplest 
rationalizing  factor  for  V5  +  VS  is  its  conjugate,  VS  —  \/3  (Prin.). 

12.  Find  the  simplest  rationalizing  factor  for  Va  +-v^. 

Solution.  — By  §  111,  Va  +  \/b^,  or  a^  +  fit,  is  exactly  contained  in  the 
sum  of  any  like  odd  powers  of  a^  and  fet,  and  also  in  the  difference  of  any 
like  even  powers  of  a^  and  ftt.  Since  in  raising  ai  and  fcl  to  the  same  power 
the  exponents  |  and  f  are  multiplied  by  the  index  of  the  power,  the  lowest 
like  powers  of  a^  and  bs  that  are  rational  numbers  are  the  sixth  powers, 
which  are  even  powers.  Hence,  the  rational  expression  of  lowest  degree  in 
which  ai  +  fit  is  exactly  contained  is  (a^)^  -  (6^)*,  or  a^  —  b*. 

Dividing  a'  —  6*  by  a^  +  bh  or  by  Va  +  y/h^,  the  rationalizing  factor  is 
found  to  be  a2  —  a^ftt  +  a^b^  -  ab^  +  a^b^  —  b^. 


RADICALS  263 

Find  the  simplest  rationalizing  factor  for 

13.  V3  +  V2.  16.    Va  +  V^.  19.    -s/a^  +  ^• 

14.  V3-V2.  17.    a-2V6.  20.    Va--^. 

15.  2  +  V3.  18.    3V^  +  2?/.  21.    y/x+-\/f. 

3 
22.    Rationalize  the  denominator  of 


Vl2 

Solution 
3    _    3x\/3    ^3\/3^3V3^V3  q,,  i^^ 
VT2      VB  X  V3      V36         6  2' 

Rationalize  the  denominators : 

23.    1-.  26.    ^.  29.    -i^.  32.      ^ 


V3  V6  ^x'  ^ax" 


24.  A.  27.    -1=.  30.        ^.  33.    V«+^. 

V5  V'4  V2  aa;  Va  —  h 

25.  J|.  28.    2^.  31.    #.  34.    ^^^H?. 

^3  V6y  ^12  a/^+2 

35.  Rationalize  the  denominator  of      _  ~      _» 

V7  +  V3 

Solution 
V7  -  V3  _  C  V7  -  >/3) (\/7  -  V3)  ^  7  -  2  V21  +  3  _  10  -  2V21  _  5  -  V2T 
V7+V3~(\/7+V3)(>/7-V3)  7-3  4  2     ^* 

Rationalize  the  denominators : 

36.  ^^  +  A  38.       ^^-^.  40.      1--^^ 


V3  -  V2  V2  +  V3  V8  -  V7 

37.    ^.  39.  ^-^^.  41.    V|±V5. 

V5  -  V3  2  -  V2  Va;  -  V^ 

42.  4V2  +  6V3  ^^     a;-V^^^=T 
3V3-2V2  '   a;  +  V^^31 

43.  V^  +  1  —  2.  ^g     aVa—  Va;  +  1 
Va;  +  1  +  2  *    Vo^  +  V^Tl 


264  RADICALS 


46.    ^x  +  y-^/x-y  ^^     \/a?  +  a-\-l  —  l 

-s/x  +  y  +  Va.  —  y  Va'^  +  a-\-l+l     '     . 

5 

48.  Find  the  approximate  value  of 

V3 

SOLCTION 

6   _  by/S  _  6  X  1.73205  _  ^  ggg^g 
V3        3  3 

In  solving  the  following  examples,  the  work  will  be  lessened  by  observing 
that  V2  =  1.41421,   V3  =  1.73205,  and  \/5  =  2.23607. 

Find  the  approximate  value  of 

49.  A..  52.    J^.  55.    8l^. 

V2  V45  2-V3 

50.  J-.  53.    ^.  56.    1±^. 
V5                                V50  2  -  V2 

Ri       6  -.         1  fi7    5  4-2V5. 

51. 54.    — ♦  o7.    — • 

V8  V125  6  -  2  V6 

58.  Simplify  — = ^ ^• 

V2+V3+V5 

Solution 
Since  (V2  +  V3)-V5  =(V2 -\/5)  + V3, 

y/2-VS-V5^(V2-V5)-VS^,  (V2-V6)  +  V3 
y/2+V3+y/l      (V2  +  \/3)  +  \/5      (\/2+\/3)-V^ 
_  2  -  2  VlO  +  5-3_4-2v/l0 
2  +  2V6  +  3-5  2v'6 

_2-\/T0_2v^-2VT5_  y/e-VIS 
~      V6     ~  6  3        * 

E-ationalize  the  denominators : 

59.  V2-V5-V7.  61 


V2+V5+V7  V2+V3+V5 

V3+V2  g2    2V2-3V3  +  4V 

V34.V2-V6'  '       V2+V3-V5 


RADICALS  265 

276.  To  find  the  square  root  of  a  binomial  surd. 

(V2  +  Ve)"  =  2  +  2  Vi2  +  6  =  8  +  2  Vl2; 
(V4  -H  V3)2  =  4  +  2  Vl2  +  3  =  7  +  2  Vl2 ; 
or  (2  +  V3)2  =  4+4V3    +3  =  7  +  4  V3. 

1.  Since  (V2  +  V6)^  =  8 +2  Vl2,  what  is  the  square  root  of 
8  +  2  Vi2  ? 

2.  How  does  the  product  of  the  terms  of  the  square  root  of 
8  +  2  Vl2  compare  with  the  irrational  term  2  Vl2  ? 

3.  How  does  the  sum  of  the  squares  of  the  terms  of  the  square 
root  compare  with  the  rational  term  8  ? 

4.  How,  then,  may  the  square  root  of  8  +  2  Vl2  be  found  from 
the  termsof  8  +  2V12? 

5.  How  may  the  terms  of  the  square  root  of  7  +  2  Vl2,  or  the 
equivalent  expression  7  +  4  V3,  be  found  ?  After  the  irrational 
term  is  divided  by  2,  what  two  factors  of  the  result  are  selected 
for  the  terms  of  the  root  ? 

277.  A  surd  of  the  second  degree  is  called  a  Quadratic  Surd. 

Vx,  4  Va;,   Vx  +  Vy,  and  3  +  2  y/b  are  quadratic  surds. 

278.  Principlk.  —  The  terms  of  the  square  root  of  a  quadratic 
binomial  siird  that  is  a  perfect  square  may  be  obtained  by  dividing 
the  irrational  term  by  2  and  then  separating  the  quotient  into  two 
factors,  the  sum  of  whose  squares  is  the  rational  term. 

Examples 
1.   Find  the  square  root  of  14  +  8  V3. 

Solution 

Since,  if  14  +  8  VS  is  the  square  of  a  binomial  quadratic  surd,  the  irra- 
tional term  8  V3  is  twice  the  product  of  the  terms  of  the  root  (Prin.),  4  \/3, 
or  V48,  IS  the  product  of  the  terms  of  the  binomial  surd.  Since  the  two 
factors  of  V48,  the  sum  of  whose  squares  is  14,  are  VQ  and  Vs,  the  required 
square  root  is  equal  to  VB  +  VS. 


.-.  Vl4  +  8  \/3  =  V6  +  V8. 

ALG.  — 17 


266  RADICALS 

2.   Find  the  square  root  of  11  —  6  V2. 

Solution 


Vll  -  6v/2  =  Vll  -  2>/l8  =(3  -  v^) 

=  3  -  \/2. 

Find  the  square  root  of  each  of  the  following : 

3.  12  +  2V35.  11.  12  +  4V6. 

4.  16-2V60.  12.  11+4V7. 

5.  15  +  2V26.  13.  12-6V3. 

6.  16-2V55.  14.  17  +  12V2. 

7.  114-2V30.  15.  15-6V6. 

8.  7-2V10.  16.  18  +  6V5. 

9.  3-2V2.  17.  a'  +  b  +  2aVb. 
10.  6-I-2V5.  18.  2a-2Va2-6'. 

PROPERTIES  OP  QUADRATIC  SURDS 

279.  The  square  root  of  a  rational  number  cannot  be  partly 
raiional  and  partly  a  quadratic  surd. 

For,  if  possible,  let  Vy  =  Vb  ±  m. 

By  squaring,  j/  =  6  ±  2  mVft  +  m', 

and  Vb=±y-'^'-^; 

2  m 

that  is,  a  surd  is  equal  to  a  rational  number,  which  is  impossible.    Therefore, 
Vy  cannot  be  equal  to  Vb  ±  m. 

280.  In  any  equation  containing  rational  numbers  and  quadratic 

surds,  as  a  -{-  y/b  =  x  -\-  -Vy,  the  rational  parts  are  equal,  and  also 
the  irrational  parts. 

Let  a  +  Vfe  =  X  +  Vy.  (1) 

Since  a  and  x  are  both  rational,  if  possible,  let 

a  =  X  ±  »n.  (2) 

Then,  x±m  +  \/6  =  x  +  Vy,  (3) 

and  Vy=Vb±m.  (4) 


RADICALS  j  267 

Since,  §  279,  equation  (4)  is  impossible,  a  =  x  ±m  is  impossible ;.  that  is, 
a  is  neither  greater  nor  less  than  x. 

Therefore,  a  =  x,   and,   Eq.  (1),  Vb=y/y. 

Hence,  if  a  +  Vb  =  x  +  Vy,  a  =  x,  and  Vb  =  VM 


281.   If  Va+V6  =Vx-\- Vy,  then  Va  —  V&  =  V'a;  —  ^/y,  when 
a,  b,  X,  and  y  are  rational  and  a  >  V6. 

For,  squaring,  a +  ^^1)  =  x +  2Vxy  +  y. 

Therefore,  §  280,        a  =  z  +  y,  and  Vb-=2  Vxy. 

Hence,  a—-\/b  =  x  +  y  —  2  Vxy. 


Whence,  v  a  —  Vb  =  Vx  —  y/y. 

Examples 
1.    Find  the  square  root  of  21  +  6ViO. 

Solution 
Let  y/x-\-Vy=:  V2I  +  6VIO.  (1) 

Then,  §  281,  \/x-Vy  =  V2I  -6v/l0.  (2) 

Multiplying,  x-y=  \/441  -  360  =  Vsl, 

or  x~y  =  9.  (3) 

Squaring  (1),  x  +  2Vxy  +  y  =  21  +  OVlO. 

Therefore,  §  280  a;  +  y  =  21.  (4) 

Solving  (4)  and  (3),  x  =  15,  ?/  =  6. 

.'.  Vx  =  W5,  Vy  =  V6. 
Hence,  the  square  root  of  21  +  6  VTo  is  -v/lS  +  V6. 

Find  the  square  root  of 

/2.  25  +  10V6.  8.  16  +  6V7.  14.  2+V3. 

3.  19  +  6V2.  9.  21-8V5.  15.  6+V35. 

4.  45  +  3OV2.  ^0.  47  -  12 ViT.  V  16.  1  +  |V2. 

5.  35-14V6.  11.  56  4-32V3.  17.  2 -I- f  VO. 

6.  11  +  6V2.  vJ2.  35  -  12 V6.  18.  30  4-  20V2. 

7.  24-8V5.  13.  56-12V3.  19.  18  -  6 V5. 


268  RADICALS 

RADICAL  EQUATIONS 

282.  An  equation  involving  an  irrational  root  of  an  unknown 
number  is  called  an  Irrational,  or  Radical  Equation. 

X*  =  3,  Va;  +  1  =  Vx  —  4  +  1,  and  y/z  —  1  =  2  are  radical  equations. 

283.  An  equation  containing  quadratic  surds  involving  x  may 
be  rationalized  with  respect  to  x  by  writing  all  the  terms  in  the 
first  member  and  multiplying  both  members  by  the  proper  ration- 
alizing factor. 

To  rationalize  Vx  —  .3  =  2.  (1) 

Transposing,  Vaj  -  3  -2  =  0.  (2) 

Multiplying  by  the  conjugate  surd, 


(Vx-3-2)(Vx~- 3  +  2)=0,  (3) 

or  X  -  3  -  4  =  0.  (4) 

The  rationalization  of  such  equations,  however,  is  more  con- 
veniently accomplished  by  the  process  of  squaring. 

Thus,  by  squaring  (1),  (4)  may  be  obtained  in  the  form 

X  -  3  =  4.  (5) 

It  is  evident,  then,  that  squaring  an  equation  is  equivalent  to 
multiplying  both  members  by  the  same  unknown  expression,  an 
operation  likely  to  introduce  roots  (§  196).  The  roots  introduced, 
if  any,  when  an  equation  is  rationalized  are  those  of  the  equation 
or  equations  formed  by  placing  each  rationalizing  factor  equal  to 
zero. 


Thus,  in  squaring  the  equation  Vx  —  3  =  2,  the  root  of  Vx  —  3  =  —  2  is 
introduced.  But  if  Vi  is  taken  to  mean  either  +  2  or  —  2,  this  equation  has 
the  same  root  as  the  given  equation  and  no  root  has  been  introduced. 

Repeated  squaring  is  often  necessary  to  free  an  equation  of 
quadratic  surds  involving  x.  This  corresponds  to  repeated  ration- 
alization with  respect  to  particular  surds. 

Thus,  Vx  -  5  =  -  Vx  +  5. 

Squaring,  x  -  10  Vx  +  25  =  +  (x  -f  6)  =  x  +  5. 

Simplifying,  y/x  =  2. 

Squaring,  x  =  4. 

Or,  transposing  in  the  given  equation, 

Vx  -  6  +  Vx  +  6  =  0, 


RA  DICALS  269 

Rationalizing  with  respect  to  Vx  +  5, 

(Vx-  5  +  \/x  +  5)(\/S-5— Vx  +  5)  =  0. 

Expanding,  x  —  lOVx  +  25  —  (x  +  5)  =  0. 

Simplifying,  Vx  —  2  =  0. 

Rationalizing,  (  Vx  —  2)  (  Vx  +  2)  =  0, 

or  X  —  4  =  0. 

In  squaring  the  first  time,  the  factor  Vx— 5— Vx+5  is  introduced;  and 
in  squaring  the  second  time  the  factor  Vx+2,  which  is  ^  of  the  product 
( Vx  +  5+  Vx  +  5)(  Vx  +  5  —  Vx  +  5),  or  10 Vx  +  20,  is  introduced. 

284.    It  follows  from  the  preceding  discussion  that : 
If  a  radical  equation  is  rationalized  by  multijylying  by  a  rational- 
izing factor  or  by  squaring,  the  resulting  equation  has  all  the  roots 
of  the  given  equation. 

Whether  the  given  radical  equation  has  all  the  roots  of  the 
rational  equation  depends  upon  the  method  agreed  upon  of 
verifying  radical  equations. 

As  illustrated  above,  each  of  the  equations 

Vx  -  5  +  Vx  +  5  =  0,  (1) 

v^  _  5  _  Vx  +  5  =  0,  (2) 

Vx  +  5  +  Vx+5  =  0,  (3) 

and  Vx  +  5  -  Vx  +  5  =  0  (4) 

is  rationalized  by  finding  the  product  of  them  all,  which  is  x— 4=0.  Hence, 
the  equation  x  —  4  =  0  has  the  roots  of  the  four  equations ;  that  is,  each 
equation  has  the  root  x  =  4,  or  has  no  root. 

If  X  =  4,  Vx  =  V4  and  Vx  +  5  =  VO.  If  Vi  is  either  +  2  or  -  2  and  V9 
is  +  3  or  —  3,  the  equations  are  verified  as  follows : 

(1)  becomes  2-5  +  3  =  0, 

(2)  becomes  2-5-(-3)=0, 

(3)  becomes  -2  +  5+(-3)=0, 

(4)  becomes  -2  +  5-3  =  0. 

To  prevent  confusion  in  making  numerical  substitutions,  it  is 
customary  to  regard  only  the  positive  or  principal  square  root  in 
expressions  like  Vi,  V9,  V5,  etc.     (See  §  225.) 

For  example,  by  common  agreement  +  V9  means  +(+3),  or  +3,  —  V9 
means  —(+3),  or  —3;  +V6  means  the  positive  square  root  of  5;  etc. 
With  this  understanding  equations  (2),  (3),  and  (4)  cannot  be  verified  for 
X  =  4,  and  since  they  have  no  other  root,  they  may  in  this  sense  be  regarded 
as  impossible  equations. 


270  RADICALS 

When  the  equations  given  in  this  section  have  been  freed  from 
the  radical  signs,  the  resulting  equations  will  be  found  to  be 
simple  equations.  Other  varieties  of  radical  equations  are  treated 
subsequently. 

Examples 

1,   Given  V2  cc  +  4  =  10,  to  find  the  value  of  x. 

Solution 
y/2z  +  4  =  10. 
Transposing,  y/2x  =  6. 

Squaring,  2  a;  =  36. 

.-.a:  =18. 


2.   Given  Va;  —  7  +  Va;  =  7,  to  find  the  value  of  x. 

Solution 

X-  7  f  yJx  =  7. 


Transposing,  Va;  —  7  =  7—  y/x. 

Squaring,  a;  —  7  =  49  —  HVx  +  X. 

Transposing  and  combining,     14  Vx  =  56. 
Dividing  by  14,  Vx  =  4. 

Squaring,  x  =  16. 


3.  Given  \14  +  Vl  +  Va;  +  8  =  4,  to  find  the  value  of  x. 


Solution 


\14+Vl+Vx  +  8  =  4. 


Squaring,  14  +  Vl  +  Vx  +  8  =  16. 


Transposing,  etc.,         Vl  +  \/x  +  8  =  16  -  14  =  2. 


Squaring,  .  1  +  Vx  +  8  =  4. 


Transposing,  etc.,  Vz  +  8  =  4  —  1  =  3. 

Squaring,  z  +  8  =  9. 

.-.  a;  =  9-8  =  l. 


Verification.  \i4  4.  Vl  +  vT+8  =  Vi4  +  Vl  +  3 

=  V14T2  =  4. 


RADICALS  271 


4.   Given  2y^-_&  = 
2  Vcw;  +  6 

:  3^  -  2  ^^  to  find  the  value  ( 
3  Voic  +  36 

Solution 

2Vax-6 

3V«x-26 

2  Vox +6 

3Vax  +  36 

2  7) 

-1             56 

2Vax  +  6 

3Vax  +  36 

Canceling, 

26 

66 

2Va«  +  6 

3\/ax  +  36 

Dividing  by  —  6, 

2 

5 

2\/ax  +  6 

3Vax  +  36 

Clearing  of  fractions,  etc. 

,,  6Vax-  10%/ax  = 

=  56-66. 

.%    a/ox: 

_6 
4 

Squaring,  etc., 

X  - 

62 
16  a 

Suggestions.  —  1.  When  the  equation  is  free  from  fractions, 
transpose  so  that  the  radical  term,  if  there  is  hut  one,  or  the  more 
complex  radical  term,  if  there  is  more  than  one,  mxiy  constitute  one 
member  of  the  equation ;  then  raise  each  member  to  a  power  of  the 
same  degree  as  that  radical.  Simplify  the  result.  If  the  equation 
is  not  freed  from  radicals  by  the  first  involution,  proceed  again  as 
at  first. 

2.  It  is  sometimes  convenient  to  rationalize  denominators  before 
clearing  of  fractions  or  involving. 

Solve  the  following  equations  : 

5.  Va;  +  11  =  4.  12.    l  +  2Vi  =  7-V». 

6.  Va;  +  5  =  3.  VlS.    Va;  +  16  -  V^  =  2. 


7.  Va;-a2=6.  1^14.  \/2»  -  V2 a;  -  15  =  1. 

8.  ^a; -  1  =  2.  15.  Var«  +  a;  +  l  =  2-a;. 
A-  \/ar^^  =  a.  16.  3V^^^  =  3a;-3. 

10.  Vx-\-h  =  a.  17.  Vaf  +  2=Va;  +  32. 

11.  1  +  V«  =  5.  18.  5  —  Va;  +  5  =  Va. 


272  RADICALS 

19.    2Vx  —  x  =  x-8Vx.  25.    V3 a; -  5  +  V3 a;  +  7  =  6. 


20.    V4a^+6a;-10=2a:+4. 


21.    Va^  — 5a;-f  7  +  2  =  a;. 


26.  Vl6a;  +  3+Vl6x4-8  =  6. 

27.  V9  ic  +  8  +  V9^  -4  =  0. 

22.  4-V4-8a;  +  9a^  =  3a;.      28.  ^1  +  a; V^^+H  =  1  +  ». 

23.  V2(l-a;)(3-2a;)-l=2a;.    29.  a/t  +  3 V5¥^^36  -4  =  0. 


24.    V2a;-l+V2»  +  4  =  5.      30.    2a;+A/4a:2- Vl6a^-7=1. 
31.    V7  +Vl  +V4  +>/l  +2 V^  =  3. 


32.  ^        =  V3a;  +  2+V3a;-l. 

V3a;  +  2 


33. 


34. 


35. 


36. 


37. 


V2x  +  9^  V2a;4-20 
V2a -  7      V2^ -  12 

V^  + 18         32 


v^  +  2      V^  +  6 


4-1. 


■y/x  —  1      Va;  — 3 


Va;  +  5      Va;  —  1 

y/x  —  6  _  -y/x  —  8 
V»  —  1      Va;  —  5 


■y/x  —  3  _  Va;  —  4 

Va;  +  1      Va;—  2 


38. 


39. 


^40. 


41. 


42. 


^  +  6     V2x  +  2 


V2a;  +  4      V2a;  +  1 


y/nic  -f  V2  a;  +  3  ^  8 
VlT^-V2a;-|-3     3 


2V2x  +  4     3  Va;  +  1  +  9 


2V2a;-4      3VST1-9 
A/V5«+11     ^iV5x-lQ 


V4a;  +  3  +  2Va;-l 

V4a;  +  3-2Vx^^ 


=  5. 


43. 


44. 


45. 


Vaj  4- 1  —  Va;  —  I      1 


Va;  +  1  +  Va;  —  I      2 

_^-3_^V^+V3_^2V3. 
V^-V3  2 

Vi9a;-f-V2a;-|-ll 


=  2*. 


Vl9a;-V2a;-Hll 
46.   2V^-V4a;-22-V2  =  0. 


RADICALS 


273 


47.  l  +  V(3-5a;)2  +  16  =  2(3-a;). 

48.  -\/x  +  yx  —  Va^  —  x=  y/a. 


49.    Va+Va;— (a  — 6)^  =  a  +  6. 


50.  Vwiw  —  a;  —  Va;  Vwn  - 

51.  aV»  —  ftVS  =  a-  +  6^ 


1  =  Vwin  Vl  —  X. 
-2  ah. 


62.    V5  ax  — 9  a'  +  a  =  VSaa;. 
6  a 


53.    -\/x  -j-  3  a  = 


V»  +  3  a 


—  v». 


54.    V2^-V2 


a; 


7  = 


V2a;-7 
55.    \/2a;  +  Vl0a;  +  l=V2^  +  l 

56. 


Va;  +  g  +  Va;  —  a  _  2  4.  Va^*^  —  «^ 
Va;  +  a  —  Va;  —  a  ** 

57.    V»  +  V2^  +  V3^  =  Va. 

Solution 
Vx  +  y/2x  +  V3«  =  Va. 
Factoring,  VicCl  +  V2  +  V3)  =  Va. 

Multiplying  by  1+  a^  -  V3, 

v^(l  +2V2  +  2-3)  =  V5(l  +V2-V3). 
Vx  •  2  V2  =  Va(l  +  V2  -  V3). 
Squaring,  8  a;  =  a  (1  +  V2  -  V3)2. 

X  =  ^  (1  +  V2  -  V3)3. 

■^58.    V2^  +  V3^  +  V5  a;  =  Vw. 
1^59.    V2^  +  V3x  —  VS^  =  Vc. 
60 


V^^^4-V2(a;-a)=  A|3a;  +  aV2. 
61.    V^^n+V2a;-2  =  V3a;-3  +  V2. 
"3  +  V4a;-6  =  V2a  +  V^. 


.^62.    V2^ 

ALG.  —  lb 


274  REVIEW 

REVIEW 

Reduce  the  following  to  their  simplest  forms : 

Qa^  —  7  0^  —  5x^  X  —  y     y  +  x      4a^^ 

9a^  —  25x  x  +  y     y  —  X     ai*  —  y* 

2    8x'  +  18x-5  g     V2-V3      V2  4-V3 

•    12a^  +  5a;-2*  "    V2-|-V3      V2-V3" 

g^    aV  —  aVx  -I-  a;^  a;+V^+y      xVx+yVy 


2  a  V6  +  6 


a 


10.    —r^ =-1  + 


2+V5      V243  1-V2a;    1  +  V2a;    l-2a; 

13. 


a;  4-  Va^  —  a^      a;  —  Vx^ 


Vic^  —  a^     a;  +  Var* 


14. 


Va  +  1  +  Va  —  1      Va  4- 1  —  Vc 


Va  +  1—  Va  —  1      Va  +  l+Va— 1 


a^-2ax-3x'       6  a'  +  7  aa;  +  2  ar' 
So? +  6  ax +  2  7?       a'-Aax  +  Saf' 

16  a'-6  ^  gg  -  4  g  V6  +  46 

d'~2aVb  +  b       a^  +  2  a  V6  +  & 

Va6 


"■  i'-tH^4}     »•  ^K^'^^/S^a 


1  +  g  +  g^  '  /  a       Va;\  /_ot Va;' 


ift     l+V«  +  a  „„     VVa;       «  7 VV^       « 


1  —  Vg  +  a  /  g        Va;\  /  a        Va;' 


1  —  Vg  vVa;       '^  /  \  Va;        * 


REVIEW  275 

Expand : 

21.  (a^-by.  25.    (a-2+a-y.  31.    (a-Vb)*. 

22.  C^a-Sby.  ''•    ^""  +  '^'-  32.    (Vx+V^)«. 

27.    (a^-6*)«.  33.    (V2-V3)*. 

^^'    (^"2)*  28.    (a*-6-i)*.  34.    (V5-2)«. 

29.    (a-*-6-i)«  35.    (^4-^)«. 

24.    (aa:  +  -j-  3^j^    (ai  +  6^)«.  36.    (V2-^/2)«. 

Extract  the  square  root  of 

37.  ^  +  3ar^-ar'-^  +  |- 

38.  ^+42/2+:^-2a^+f-.V«. 
4  lb  4 

39.  a-4-12aV  +  54a&  +  108aM  +  816* 

40.  1  +  2-y/x  —  x  —  2  xVx  +  of. 

41.  a  +  46  +  9c  — 4Va6  +  6Vac  — I2V6C. 

42.  cc^  —  4  xVxy  +  6xy  —  4:  yVxy  +  3/^ 

43.  a^-12  aV  +  60  x^y  -  160  a;?/^  +  240  tcV 

-  192  ic*?/2  +  64  f. 
Find  the  square  root  of 

44.  81234169.  48.   56  +  14 Vl5. 

45.  64064016.  49.   47  -  12 Vi5. 

46.  .00022801.  50.   62  +  20V6. 

47.  .1  to  four  places.  51.   51  -  36V2. 

Extract  the  cube  root  of 

52.    a:3_9a.+  27a;-i-27a;-3. 

64.   cc^  -f-  3  ar* V^  —  5  xVx  -f  3 Va;  —  1. 


276  REVIEW 

55.  Find  the  cube  root  of  2  V2  -  6^2  +  3  V2  ^/4  -  2. 

56.  Find  the  cube  root  of  (a  +  12  &  +  3  c)Va 

-(6a  +  86  +  6  c)V6  -  (3  a  +  12  &  +  c)Vc  +  12Va6c. 

57.  Extract  the  cube  root  of  510,082,399. 

58.  Extract  the  cube  root  of  1,042,590,744. 

59.  Extract  the  cube  root  of  2  to  three  decimal  places. 

60.  Find  the  first  four  terms  of  VI  +x  —  x'. 

61.  Find  the  first  three  terms  of  ^1  +  a^. 

62.  Find  the  fourth  root  of 

a«  -  4  a^VoFi  +  6  a^b-^  -  4  ab-'Vab^'  +  6-^ 

63.  Find  the  sixth  root  of 

8  -  48 Va  + 120  a  -  160  aVa  +  120  a^  -  48  aWa  +  8  a". 

If  a"  X  a"  =  «"*+"  for  all  values  of  m  and  n,  show  that 

64.  a-  =  l.  67.    (a&)o  =  l. 


65.    J  =  V^^  =  {Va)\  ^^-    ^«^'>'  = 


69.    f^]=: 


66.    2a-i  =  ?i!?. 

a  V& 

Find  the  values  of  the  following : 

70.  16^  73.    (a*x*)i.  76.    Qi^)'^. 

71.  27*.  74.    (byyK  77.    (36)-f 

79.  For  what  values  of  n  is  (a  —  6)"  =  (6  —  a)"  ? 
Simplify  and  express  with  positive  exponents : 

80.  (36  a-3  ^  25  a-T2.  83.    (VaFx^^ -h^/a^^i 

81.  (8  a»a*  X  64  a-*a;-T*  84.    (Va^*^  Va26)-i 

82.  (ah^y  ^  (ah^y.  85.    (Va-;-va)^\/a. 


REVIEW 


277 


86 


a  +  b        2  a^b 


a^  —  b^     a^  +  b^     a^  —  b^ 

1  +  a-'b  .  /I  +  ab-'  +  a'b-'  ^  1  +  a  ^&^ 
•    1  _  a-'b  '  [l  -  ab-'  +  d'b-^     1  -  a'^b'^j 


Solve  the  following  equations : 

a;  +  l_a7  —  1_3  —  5a; 
x—1      x  +  1       1  —  a^ 


88. 


fto    7-2a;     2^-1  ,,_  5^-61     17  +  3a; 
89.    -^^  5"  +  ^"^"  "• 


2  a; 


30 


90. 


91. 


92. 


4  a; -17      3|-22a; 
9  33 

3x  —  5y     2x 


x" 


=^-s^-s- 


KM+H 


8.v-9^y  I    7 
12  2     12 

lP^a;-^-24U0. 
3  V  8 


93.    j 

I  (a; 


3a;  +  l=2y, 

(a:  +  5)0/  +  7)  =  (a;+l)(i/-9)  +  112. 

r  a;  -  2/  =  3, 
(a;  +  l)(a;  +  2)-(a;-2)(a;+l)  =  ll2/  +  2. 


Simplify  and  express  with  positive  exponents : 


94. 


96. 


96. 


27 '  sy 

3  a-*  +  2  a; 


\a-\a\a^^Y\K 


a.-2«/j 


\xy' 


b 


97^    fa;  ^  ^  .    a;-yT 
La;-*y-i  "  (a^)"'J 

98.    Ka^&^)^-^(a"^"^l*- 

g  4-  &         a  —  b 

a*  _  &J      a3  +  6^ 

b  +  a 
ab 


99 


100. 


QUADRATIC   EQUATIONS 


285.  1.  What  is  the  value  of  x  in  the  equation  3cc  =  24? 
What  kind  of. an  equation  is  it? 

2.  What  powers  of  x  are  found  in  the  equation  a;-  +  2  x  =  3  ? 
Which  is  the  higher  power  ? 

3.  What  is  the  value  of  x  in  the  equation  a;-  =  9  ?  How  many- 
values  has  X  ?     How  do  they  compare  numerically  ? 

4.  Factor  a^  —  5a;  +  6  =  0,  and  so  find  the  values  of  x.  How 
many  values  has  x  ? 

286.  An  integral  equation  that  contains  the  square  of  the 
unknown  number,  but  no  higher  power,  is  called  a  Quadratic 
Equation,*  or  an  equation  of  the  second  degree. 

It  is  evident,  therefore,  that  quadratic  equations  may  be  of 
two  kinds  —  those  which  contain  only  the  second  power  of  the 
unknown  number,  and  those  which  contain  both  the  second  and 
first  powers. 

x^  =  15  and  aoii^  +  bx  =  c  are  quadratic  equations. 

PURE  QUADRATICS 

287.  An  equation  that  contains  only  the  second  power  of  the 
unknown  number  is  called  a  Pure  Quadratic. 

ax^  =  b  and  ax^  —  cx^  =  be  are  pure  quadratics. 

Pure  Quadratics  are  also  called  Incomplete  Quadratics,  because 
they  lack  the  first  power  of  the  unknown  number. 

288.  Since  pure  quadratics  contain  only  the  second  power  of 
the  unknown  number,  they  may  be  reduced  to  the  general  form 
axi^  =  b,  in  which  a  represents  the  coefficient  of  x^,  and  b  the  sum 
of  the  terms  that  do  not  involve  a^. 

278 


QUADRATIC  EQUATIONS  279 

289.  Principles.  —  1.  If  the  square  root  of  each  member  of  a 
quadratic  equation  is  extracted  and  the  second  member  of  the  result- 
imj  equation  is  given  the  sign  ± ,  the  resulting  equation  is  equivalent 
to  the  given  equation. 

2.  Every  pure  quadratic  equation  has  two  roots  numerically  equal, 
but  having  opposite  signs. 

For  the  equation  A-  =  B^,  or  A^  -  B^  =  0,  or  (A  -  B)(A  +  B)=  0,  is 
equivalent  to  the  two  equations 

A-  B  =  0  and  A  +  B  =  0, 

or  A  =+  B  and  A  =  —  B; 

that  is,  to  the  two  equations  A  =  ±  B. 

Examples 

1.  Given  10  XT  =  99  —  or,  to  find  the  value  of  x. 

Solution 
10x2  =  99-x2. 
Transposing,  etc.,  11  x^  =  99. 

Dividing  by  11,  x^  =  9. 

Extracting  the  square  root  of  each  member.  Ax.  7, 

x  =  ±3. 

Strictly  speaking,  the  last  equation  should  be  ±  x  =  ±  3,  which  stands  for 
the  equations  +x=+3,  +x  =  —  3,  — x=:  —  3,  and  —  x  =  +  3.  But  since 
the  last  two  equations  may  be  derived  from  the  first  two,  the  first  two  express 
all  the  values  of  x.  For  convenience,  the  two  expressions,  x  =  +  3  and 
X  =  —  3,  are  written  x  =  ±  3. 

Consequently,  in  extracting  the  square  roots  of  the  members  of  an  equa- 
tion, it  will  be  sufficient  to  write  the  ambiguous  sign  before  the  root  of  one 
member. 

2.  Find  the  roots  of  the  equation  3  a^  =  — 15. 

Solution 
3x2  =  -15. 

Dividing  by  3,  x^  =  —  5. 

Extracting  the  square  root,        x  =  ±  V  —  6. 


280  QUADRATIC  EQUATIONS 

Solve  the  following  equations  : 

3.    7«^-25  =  5a^  +  73.  14.  -^ ^^^  =  0. 

a-\-h         X 

5.  {a-xy={Sx+a){x-a).      ^^'  ^^2 "*" ^TJI^ " ^^• 

6.  aa:2=(a-6)(a2-62)-6ar'.  «     a;     a6 

lb.  — I —  =  — • 

7.  a^x^+2ax'={a^-iy-a?.  x     a      x 

8.  (a;  +  2)2-4(x  +  2)  =  4.        17.  V^F+8 -^ =  x. 

10.    -^: 1-     ^     =  2|.  19  jg+g  .  a;  —  q^g^'  +  ft" 

1-a;     l  +  a;  •  x+b     x-b     x"  -  b^' 

, ,      a;    ,  a^  — 15     x  ^^ — -  6 

12         5  a;         5 


12.   ^±-?  +  £zi3  =  4_  21.  ^^       -Var'  +  12  =  a;. 

a:-3x  +  3  Var^  + 12 

13     ''^~2 _i_ ^  +  2 _ _ ;[  22     ^"^^  ,  a;  —  a        2a 


x+1     x—1  x—a     x+a     1—a 

x+7         x—1  7 


23. 


a^-7a;     a^  +  7x     x^-7S 


a' 


24.  X  +  Var^  —  d^-       

Va^  —  a? 

25.  V25  -  6a; +  V25  +  6aj  =  8. 

26.  2^±V^^^4 
2a;-V4ar^-l 


27.       Vl+a;     _^_     Vl  -  a;     ^^ 
1  4-  VH-  a;     1  +  Vl  — a; 

V^Ti+V^^^    '^ 

1  11 


28. 


29. 


1  +  Vl  -  »      VI  +  a;  +  1      » 


QUADRATIC   EQUATIONS  281 


2  2 

30.    -I =a?. 

X  +  V2  —  a^     a:  —  V2  —  or^ 


31. 


Vo;  -f  2  g  —  Va;  —  2  a  _  x 
Va;-2a  +  Va;  +  2a     2  a 


32.    J^^=^+\/^^  =  - 
^a;  +  a      ^a;  —  a 


Problems 

1.  If  25  is  added  to  the  square  of  a  certain  number,  the  sum 
is  equal  to  the  square  of  13.     What  is  the  number  ? 

2.  What  number  is  that  whose  square  is  equal  to  the  differ- 
ence of  the  squares  of  25  and  20  ? 

3.  If  a  certain  number  is  increased  by  5  and  also  decreased 
by  5,  the  product  of  these  results  will  be  75.  What  is  the 
number  ? 

4".  Two  numbers  are  as  3  to  4,  and  the  sum  of  their  squares  is 
equal  to  the  square  of  15.     What  are  the  numbers  ? 

5.  Two  numbers  are  as  4  to  3,  and  the  sum  of  their  squares  is 
400.     What  are  the  numbers  ? 

6.  A  gentleman  has  two  square  rooms  whose  sides  are  as 
2  to  3.  He  finds  that  it  takes  9  square  yards  more  than  twice  as 
much  carpeting  for  the  larger  room  as  for  the  smaller.  What  is 
the  length  of  a  side  of  each  room  ? 

7.  A  man  who  owns  a  field  80  rods  square  sells  one  fourth  of  it. 
If  the  part  he  sells  is  also  a  square,  how  long  is  each  of  its  sides  ? 

8.  A  man  had  a  rectangular  field  whose  width  was  f  of  its 
length.  He  built  a  fence  across  it  so  that  one  of  the  two  parts 
thus  formed  was  a  square.  If  the  square  field  contained  10  acres, 
what  were  the  dimensions  of  the  original  field  ? 

9.  How  many  rods  of  fence  will  inclose  a  square  gardei 
whose  area  is  2^  acres  ? 

10.  The  sum  of  two  numbers  is  10,  and  their  product  is  21 
What  are  the  numbers  ? 

Suggestion.  —  Represent  the  numbers  by  6  +  «  and  5  —  x. 


282  QUADRATIC  EQUATIONS 

11.  The  sum  of  two  numbers  is  16,  and  their  product  is  55. 
What  are  the  numbers  ? 

12.  The  sum  of  two  numbers  is  26,  and  their  product  is  69. 
What  are  the  numbers  ? 

13.  The  sum  of  two  numbers  is  5,  and  their  product  is  — 14. 
What  are  the  numbers  ? 

14.  Factor  a^  +  lTa+eO  by  the  method  suggested  in  the 
preceding  problems. 

Suggestion.  +  60  is  the  product  of  the  arithmetical  terms,  and  +  17  is 
their  algebraic  sum. 

15.  Separate  a?  +  2  a  — 2  into  two  factors. 

16.  Separate  or^  —  2  a;  —  1  into  tv/o  factors. 

17.  Divide  24  into  two  parts  whose  product  is  143. 

18.  The  length  of  a  ten-acre  field  was  4  times  its  width.  What 
were  its  dimensions  ? 

19.  The  sum  of  the  squares  of  two  numbers  is  394,  and  the 
difference  of  their  squares  is  56.     What  are  the  numbers  ? 

20.  A  man  has  two  square  fields  that  together  contain  51^ 
acres.  If  the  side  of  one  is  as  much  longer  than  50  rods  as  that 
of  the  other  is  shorter  than  50  rods,  what  are  the  dimensions  of 
each  field  ? 

AFFECTED   QUADRATICS 

290.  A  quadratic  equation  that  contains  both  the  second  and 
the  first  powers  of  one  unknown  number  is  called  an  Affected 
Quadratic. 

a;2  +  3  X  =  10,  4  x^  _  x  +  1  =  0,  and  ax^+bx-\-  c  =  0  are  affected  quadratics. 

Affected  Quadratics  are  also  called  Complete  Quadratics. 

291.  Since  affected  quadratic  equations  contain  both  the  second 
and  first  powers  of  the  unknown  number,  they  may  always  be 
reduced  to  the  general  form  of  ax^  -|-  6a;  +  c  =  0,  in  which  a,  6, 
and  c  may  represent  any  numbers  whatever. 

The  term  c  is  sometimes  called  the  absolute  term. 


QUADRATIC   EQUATIONS  283 

292.    To  solve  affected  quadratics  by  factoring. 

Reduce  the  equations  to  the  form  ay?  +  6a;  +  c  =  0,  and  solve  by 
the  methods  of  §  141. 

Solve  the  following  equations  by  factoring : 

1.  i»2_5a;  +  6  =  0.  7.  10a;2_27a;  +  5  =  o. 

2.  x'-5x  =  24:.  8.  Q{x'  +  l)=^lSx. 

3.  ar' -  1  =  3(a;  + 1).  9.  x^  —  {a  —  h)x  =  ab. 

4.  2a;2-7a;  +  3  =  0.  10.  2  ^ -^  ax-2  a^  =  i). 

5.  2  0^^-2; -3  =  0.  11.  d{h^  +  a?)=\Ohx. 

6.  3iK2_2a;_8  =  0.  12.  x^-2ax  +  {a  +  l){a-l)=0. 

13.  Solve  the  equation  x^  + 100  x  +  2491  =  0. 

Solution 
Since,  §  99,  100  is  tlie  sum  of  the  arithmetical  terms  of  the  factors  of 
«2  +  100  s;  4-  2491,  and  their  product  is  2491,  50  +  p  and  50  —p  may  repre- 
sent the  two  factors  of  2491  whose  sum  is  100. 
Then,  (50  +  p)  (50  -p)=  2491. 

Expanding,  2600-^2  =  2491. 

Transposing,  etc.,  p2  _  9. 

p  =  ±  3. 
50+p  =  53,   50'-p  =  47. 
Therefore,  §  130,  x^  +  100  a;  +  2491  =  (a;  +  53)  (x  +  47)  =  0, 

and  a;  =  —  53  or  —  47. 

Since  p  =  —  3  gives  no  new  values  of  50  +  p  and  50  -  p^  the  negative  value 
of  p  may  be  disregarded. 

14.  Solve  the  equation  a^  _|_  3  a;  _  208  =  0. 


x2  -f  3  a;  - 

208  = 

:0. 

Let 

(i+p)(f 

-p)  = 

:  -  208. 

Expanding, 

1 

-Jj2  = 

:  -  208. 

Solving, 

p  = 

:±¥. 

Factormg  the  given  equation. 

=  16,    f- 

p  =  - 

13. 

(x  +  16)(a:- 

-13)  = 

:0. 

'  X  = 

:  -  1«  or 

+  13. 

284  QUADRATIC    EQUATIONS 

Solve  the  following  equations : 

15.  a^  +  10a;  +  21  =  0.  28.  x^^  a; -756  =  0. 

16.  ar2  +  12a;-28  =  0.  29.  a;^- a; -506  =  0. 

17.  a^- 20  a; +  51  =  0.  30.  ar^  +  2  a;- 168  =  0. 

18.  x^^  60  a; +  891  =  0.  31.  a;=^  +  6  a;  - 135  =  0. 

19.  a;2  -  44  a;  +  403  =  0.  32.  x^  +  3  a;- 154  =  0. 

20.  a^  + 20 a; -629  =  0.  33.  a^  +  5a;  +  2  =  0. 

21.  a.2  -  30  .'>;  -  2275  =  0.  34.  a;2  +  a;-10  =  0. 

22.  a;2  +  24 a;  +  119  =  0.  35.  a^  -  x- 1  =  0. 

23.  a^  + 2 a; -323  =  0.  36.  x^- 2a;- 4^0. 

24.  a;2_ga;-475  =  0.  37.  ar^- 3a; -9  =  0. 

25.  ar^  +  8a;-768  =  0.  38.  a;2  +  4a;  +  8  =  0. 

26.  a;2  +  3a;-418  =  0.  39.  x^-\-6x-]-U  =  0. 

27.  a;2  4.  5  a;  _  546  =  0.  40.  a;2  +  8a;  =  -25. 

293.   First  method  of  completing  the  square. 

1.  What  is  the  square  of  a;  +  3  ?   ofa;  +  5?   of  a; +  10? 

2.  If  a;^  +  20  a;  are  the  first  two  terms  of  the  square  of  a  bino- 
mial, what  is  the  first  term  of  the  binomial  ? 

3.  Since  20  a;  is  twice  the  product  of  the  two  terms  of  the 
binomial,  and  the  first  term  of  the  binomial  is  x,  how  may  the 
second  term  of  the  binomial  be  found  ? 

4.  Since  the  second  term  of  the  binomial  is  10,  what  must  be 
added  to  a^-^  +  20  a;  to  complete  the  square  of  the  binomial  ? 

5.  What  term  must  be  added  to  a;^  +  6a;  to  complete  the 
square  of  some  binomial  ?     How  is  the  term  found  ? 

6.  What  term  must  be  added  to  a;^  +  8a;  to  complete  the 
square  ?    to  ar^  —  10  a;  ?    to  a;^  —  14  a;  ? 

7.  What  must  be  added  to  both  members  of  the  equation 
a;*  —  12  a;  =  13  to  make  the  first  member  a  perfect  square  ? 


QUADRATIC   EQUATIONS  285 

Examples 

1 .  Solve  the  equation  a^  —  6  a?  =  40. 

PROCESS  Explanation.  —  Completing   the    square   in   the 

first  member  by  adding  to  each  member  the  square 
0^  -  6  a;  =  40        ^f  ^alf  the  coefficient  of  a;,  x^  -  6  a:  +  9  =  49. 
a^— 6a;  +  9  =  49  Extracting    the    square    root   of    each   member, 

a;-3  =  ±7      a:-3=±7. 

Using  first  the  upper  sign  of  ±  7,  the  simple  equa- 

^  =  +'+<J^10        tion  X  —  3  =  +  7  gives  x  =  10.     Next  using  the  lower 

a;  =  —  7-j-3  =  _4      sign  of  ±  7,  the  simple  equation  x  —  3  =  —  7  gives 

X  =  —  4. 

Since  each  of  the  values  10  and  —  4  satisfies  the  given  equation  when 

substituted  for  x,  x  =  10  or  —  4. 

2.  Solve  the  equation  a;^  —  5  x  =  14. 

Solution 

x2-5x  =  14. 
Completing  the  square,  x^  —  5x  +  (|)2  =  14  +  ^  =  5^. 

Extracting  the  square  root,  a;  —  f  =  ±  f . 

Taking  the  upper  sign  x  =  f  +  f  =  7. 

Taking  the  lower  sign,  x  =  f  —  |=  —  2. 

3.  Solve  the  equation  10  a:^  +  19  a;  =  15. 

Solution 

10x2  +  19x=  15. 
Dividing  by  coefficient  of  x^  x^  +  ]^x=  \^. 

Completing  the  square,  x''  +  ^x+  (ig)2  =  j^  +  |f^  =  |^. 

Extracting  the  square  root,  a;  +  ^  =  ±  f^. 

Taking  the  upper  sign,  x  =  -  ^|  +  f^  =  f. 

Taking  the  lower  sign,  ^  =■  —  \i  —  ih  —  ~h 

Rule.  —  Transpose  so  that  the  terms  containing  x?  and  x  are  in 
one  member  of  the  equation  and  the  known  terms  in  the  other,  and 
make  the  coefficient  of  x^  unity  by  dividing  both  members  of  the 
equation  by  the  coefficient  of  a^. 

Add  to  each  member  of  the  equation  the  square  of  half  the  coeffi- 
cient of  X,  and  extract  the  square  root  of  each  member. 

Solve  the  two  simple  equations  thus  obtained. 


286  QUADRATIC   EQUATTONS 

Find  the  values  of  x  in  the  following  equations ; 

4.  x'~2x=U3. 

5.  x^  +  2x=im. 

6.  a^- 4  a;  =117. 

7.  a^- 6  a;  =  160. 

8.  a;2-8a:=180. 

9.  a;2  +  2a;  =  120. 

10.  a^  +  6a;  =  187. 

11.  a;2-12a;  =  189. 

12.  a.-2  4- 10  X  =  171. 

13.  a^- 22  a;  =  48. 

14.  a^  +  30a;=31. 

26. 


27. 


28. 

a;  +  2         X  2x 

29.    a^  +  (m  +  n)  (m  —  n)  =  2  mx. 

294.    Other  methods  of  completing  the  square. 

By  the  previous  method,  when  x^  had  a  coefficient,  the  equation 
was  divided  by  that  coefficient  so  that  the  ter^n  containing  ar  might 
always  be  a  perfect  square.  The  same  result  may  be  secured  in 
other  ways. 

Thus,  if  the  term  containing  x^  is  3  x?,  it  may  be  made  a  perfect 
square  by  multiplying  by  3 ;  if  8  x^,  by  multiplying  by  2 ;  if  ax^, 
by  multiplying  by  a. 

In  the  completed  square  a^x^  +  2  ahx  +  h"^,  it  is  evident  that  the 
third  term,  h^,  is  the  square  of  the  quotient  obtained  by  dividing 
the  second  term  by  twice  the  square  root  of  the  first. 


15. 

a^  +  3a;  =  10. 

16 

ar'-3a!  =  180. 

17. 

a^  +  15  a;  =  54. 

18. 

x'-x=  930. 

19. 

a^  +  13  a;  =  140. 

20. 

a^  -  11  a;  +  28  =  0. 

21. 

hx'-3x-2  =  (i. 

22. 

6a^-5a;-6  =  0. 

23. 

2a;2  +  9a;  =  35. 

24. 

3  a^  -  7  X  =  10. 

25. 

4a.-2-19a;  =  5. 

1            3     _10 

«  -1-  1      a;  -  1       3 

x"         3  a;  -  5 

_a;  +  2 

a;-2           2 

5 

1          a;-2_a;-7 

QUADRATIC   EQUATIONS  287 

Examples 

1.  Solve  the  equation  5  a^  +  12  a;  =  9. 

Solution 
5  a;2  +  12  a;  =  9. 
Multiplying  by  5,  25  x^  +  60  «  =  45. 

Completing  the  square,  25  x2  +  60  a;  +  36  =  81. 
Extracting  the  square  root,  5  x  +  6  =  ±  9. 

5  a;  =  -  6  ±  9. 
.-.  a;  =  f  or  -  3 

Explanation.  — Since  the  coefficient  of  x^  is  not  a  perfect  square,  it  may 
be  made  a  perfect  square  by  multiplying  the  members  of  the  equation  by  5. 

Since,  if  to  25  x-  +  60  x  there  were  added  such  a  term  as  would  make  the 
trinomial  a  perfect  square,  60  x  would  be  equal  to  twice  the  product  of  the 
square  root  of  25  x^  and  the  square  root  of  this  third  term,  the  square  root 
of  the  third  term  is  obtained  by  dividing  60  x  by  2  -\/25  x^  ;  that  is,  by  10  x. 
60  X  H-  10  X  =  6,  and  &^,  or  36,  added  to  both  members  completes  the  square 
of  (5  X  +  6). 

2.  Solve  the  equation  8a^  —  10x  =  3. 

Solution 
8x2-10x  =  3. 
Multiplying  by  2,  16  x*  -  20  x  =  6. 

Completing  the  square,  16  x^  -  20  x  +  (|)2  =  6  +  ^  =  ^. 
Extracting  the  square  root,  4  x  —  |  =  ±  |. 

4x  =  |±|  =  6  or  -1. 
.-.  X  =  f  or  —  \. 

General  Rule.  —  Transpose  so  that  the  terms  containing  x^  and  x 
are  in  one  member  of  the  equation  and  the  known  terms  in  the  other. 

If  the  term  containing  the  second  power  of  the  unknown  number  is 
not  a  perfect  square,  make  it  such  by  multiplying  or  dividing  the 
members  of  the  equation  by  some  number. 

Add  to  each  member  of  the  equation  the  square  of  the  quotient 
detained  by  dividing  the  term  containing  the  first  power  of  the 
unknown  number  by  twice  the  square  root  of  the  term,  containing  the 
second  power. 

Extract  the  square  root  of  each  member,  and  solve  the  two  simple 
equations. 


288  QUADRATIC  EQUATIONS 


Solve  the  following  equations : 

3.    2a^-5x  =  4:2. 

8.   3aj2  +  4cc  =  95. 

4.   6ar'-5a;  +  l  =  0. 

9.    7a^4-2a;  =  32. 

5.   4a^-12a;  =  27. 

10.    8a^-18ic  =  5. 

6.   83^^  +  200;  =  48. 

11.    6a^  +  5a;  =  4. 

7.    lSx^  +  6x  =  4:. 

12.    5a^  +  6x  =  S. 

13.   Solve  the  equation 

ax^  +  bx  +  c  =  0. 
Solution 

ax2  +  bx  +  c  =  0. 

(1.^ 

Transposing  c, 

ax^  +  bx  =  —  c. 

(21 

Multiplying  by  a, 

a'^x'^  +  abx  =  —  ac. 

(3) 

Completing  the  square, 

7)2     ;)2 
a^x^  +  abx  +  —  =  ~-ac. 
4       4 

(4) 

Multiplying  by  4, 

4  a^x^  +  iabx  +  b'^  -  b^  -  4  ac. 

(5) 

Extracting  the  square  root, 

2ax  +  b  =  ±  Vft^  -  4  ac. 

(6) 

.  „      -  6  ±  V62  -  4  ac 

(7) 

2a 

It  is  evident  that  (5)  can  be  obtained  by  multiplying  (2)  by  4  a  and  add- 
ing b^  to  both  members.  Hence,  when  a  quadratic  has  the  general  form  of 
(1),  if  the  absolute  term  is  transposed  to  the  second  member,  as  in  (2),  the 
square  may  be  completed  and  fractions  avoided  by  multiplying  by  4  times  the 
coefficient  ofx^  and  adding  to  each  member  the  square  of  the  coefficient  ofx  in 
the  given  equation. 

This  is  called  the  Hindoo  method  of  completing  the  square. 

Solve  the  following  equations  by  the  Hindoo  method: 


14. 

2x2  +  3a;  =  27. 

21. 

4a;2-a;-3  =  0. 

15. 

2a^  +  5x  =  l. 

22. 

5a^_2a;-16  =  0. 

16. 

2x'  +  lx  =  -Q. 

23. 

3a^  +  7a;_  110  =  0. 

17. 

3  ar^  -  5  a;  =  2.  " 

24. 

2cc2-5a;-150  =  0. 

18. 

4ar'-15aj  =  4. 

25. 

3a^  +  a;-200  =  0. 

19. 

5ot?-lx  =  -2. 

26. 

15a^_7a;-2  =  0. 

20. 

6a^  +  5a;  =  -l. 

27. 

7ar'-20a;-32  =  0. 

QUADRATIC  EQUATIONS  289 

295.   To  solve  quadratics  by  a  formula. 

Since  every  quadratic  can  be  reduced  to  the  general  form 
aoi?  -f  6a;  +  c  =  0,  in  which  a,  b,  and  c  represent  any  numbers 
whatever,  and  since  the  roots  of  this  equation  are 


Ex.  13,  §  294,  X  =  -&±V&^-4ac^ 

2a 

the  vahies  of  the  unknown  number  in  any  affected  quadratic 
equation  may  be  found  by  substituting  the  coefficient  of  or^  for  a, 
the  coefficient  of  x  for  6,  and  the  absolute  term  for  c. 

Examples 

1.  Solve  the  equation  6  ar*  —  x  — 15  =  0, 

Solution.  — Since  a  =  6,  6  =  —  1,  and  c  =  —  15,  by  the  above  formula, 

^^l±V(-l)-^-4  x6(-^15) 
2x6 

12  3  2 

Solve  by  the  above  formula : 

2.  2x'-ir5x  +  2  =  0.  10.  2ar'  +  3a;-l  =  0. 

3.  3ar^  +  llx  +  6  =  0.  11.  ^^  +  2x-4.  =  0. 

4.  Q3?-lx-\-2  =  {).  12.  a2_5a;  =  _3. 

5.  4r'  +  4a;-15  =  0.  13.  S7?-&x  =  -2. 

6.  2«*  +  3a;-9  =  0.  14.  4ar'- 3a:- 2  =  0. 

7.  2x2  +  3a;  +  l=0.  15.  ar'- 6a;  +  10  =  0. 

8.  3a;*-13a;  =  10.  16.  a?  +  4.x  +  12  =  0. 

9.  7a;^  +  9a;  =  10.  17.  ar«-8a;  =  -20. 

Solve  by  any  method : 

18.  a;2_6a;4-5  =  0.  22.  a*-12x  =  64. 

19.  ar»-8a;  +  7  =  0.  23.  18  ar' +  6  a;  =  4. 

20.  2a;2-5a;  =  42.  24.  ar^-a;-72  =  0. 

21.  7ar'  +  2x  =  32.  25.  4 ar* -  12 a;  =  27. 

ALG.  —  19 


290  QUADRATIC  EQUATIONS 

26.  x'  =  Sx  +  10.  gg     1  +  x     x-1 

27.  a^- 30  =  13a;. 

28.  a:2_i2a;  =  28.  39. 

29.  a^-50x-=159. 

30.  a^  +  8a;  =  84. 


33. 


9(x  - 1)         6 


40. 


31.  x  +  ^-^  =  0.  "• 

X     2 

32.  ^-H?=:?^.  42. 
9      3      4 


a^-2.  43. 


a; 

-3 

a;—  2 

5 

X 

a;  —  5 
a; 

3 

X 

-5 

2 

X 
X 

a;  4- 12 
x  +  ij 

=  7. 

X 

X 

+  4  ,  ^     x  +  3 
-2'  ^-x-3' 

4  a; 

a;  +  3 

-  4 

X 

-1 

X 

X 

+  2^ 

1  a;  +  2 

2  2  a; 

X 

5  a; 

a;  +  6 
a;-f3 

=  3. 

X 
X 

+  2 

-7 

a;  +  5 
a;  —  5 

=  1. 

X 
X 

tI- 

a;  +  2 
x-2 

23 
"lO" 

2 

a;  +  l 

5     X 

-8 

1 

-2a; 

7 

2 

34. =-.  44. 

a^-2x  +  l     4 

35.  ^_?_?^28.  45. 
4       3 

36.  —-^  +  —^  =  4.  46. 
2x  +  l      aJ-3 

37.  '^^-1=^^.  47. 

a;-|-2       x-2 


Literal  Equations 

1.    Solve  the  equation  a;''  —  -a; a;  +  l  =  0. 

b         a 


'  Solution 

a;2_«a;-^a;  +  l=0. 
b        a 

Factoring,  fx-'^(x--\  =  0. 

Therefore,  x  =  "  or  -. 

b        a 


QUADRATIC  EQUATIONS  291 

Solve  by  any  of  the  preceding  methods : 
2.    x^  —  ax  =  ah  —  hx.  Q.    ^x  —  2ax  =  3?  —  \0a. 

Z.    <x?-\-ax  =  ac  +  cx.  7.    a:^  +  3 6a;  =  5  ccc  + 15  6c. 

4.  ar^  =  (m  —  n)  a;  +  mn.  8.    2  a6a;  —  a^  =  14  a6  —  7  a;. 

5.  ar'-3  6a;=2ttaj  — 6a6.  9.    Qx^-ir^ax  =  2hx  +  ah. 

10.  oca?  —  6ca;  —  6d  -f  adx  =  0. 

11.  a:^ +  4  ma;  +  3  7ia;  + 12  mn  =  0. 

12.  x^  —  2ax  =  a^.  „_     x  .  a     5 

Ab.    --\ -• 

13.  ar'  +  4  6a;  =  6l  a      x     ^ 

14.  a;2  =  4aa,_2a2.  27.   ^  +  ^  =  1- 

4  3        3 

15.  x^  —  ax  +  a^  =  0. 

28     _1^__^  =  3 

16.  ^2  =  6a; +  61  '    a;-a     a;  +  a 

17.  a;2^^^^^^0.  ^  2a 

29. ^  — • 

18.  a;'-2a;  +  a  =  0.  a;-a  a; 

19.  4aa;-a;2^3(ij2  ^^         1      ^^      aa;  -  4 

aa;  +  4  16 

20.  ar  — a  =  l  +  aa;. 

21.  x"  -  b' =  a"  -  bx.  31.    ar'  +  ^a;  =  ^^. 

6  6 

22.  2162-4  6a;  =  a;2. 

„„       „  ,  o      (2a2  +  l)a; 

23.  5aa;  +  6a^  =  6ar^.  ^2.    ^  +  2  =  ^       ^     ^   ■ 

24.  a;^-l  =  4aa;-al  ^^     ^      2a;^4(a6-l) 

25.  a;2  4- 6^  =  4(a2  +  6a;).  *  a6  a6 

34.  a.-^  —  2 (a  —  6)a;  =  4 a6. 

35.  a;2  +  2(a  +  8)a;  =  -32a. 

36.  a;^  +  a;  +  6a;  +  6  =  ax  +  a. 

37.  2aa;  — a+2  6a;  — 6  =  2a;2  — a;. 

38.  a^ +  A(a  —  l)x  =  8a  — 4:0^. 

39.  a(a;  —  2  a  +  6)  +  a(a;  +  a  —  6)  =  a;^  —  (a  —  6)1 


292  QUADRATIC  EQUATIONS 

4n     ^'^  +  ^'  .  a  —  2  a;  _  8 


41. 


2a  —  a;     a  +  2x     3 

1  1      ^3  +  a^ 

a  —  a;      a  +  a;      a?  —  x^ 


42.    -1_+      1  1 


43. 


a;  +  a     aj  +  d      a  —  6 
a^  +  1      a  +  h  c 


X  c         a  +  b 


..     2x  —  a  ,   o         4a 
44. \- S  = 


b  2x 


45.  6ig      I  ^^a(x  +  2b) 
a  —  X  a  +  6 

46.  Va  +  a;  —  Va  —  .a;  =  V2  x. 


47.  Va;  —  a  +  V^  —  x  =  Vft  —  a. 

48.  Va;^-  6-  =  Va;  +  6  Va  +  6. 


49.    Va  —  X  +  V6  —  a;  =  Va  +  6  —  2  a;. 
50.    Solve  and  verify  Va;  +  1  +  Va;  —  2  —  V2  a;  —  5  =  0. 

Solution 

Vx+1  +  V'a;-2  -  V2x-  6  =  0. 


Transposing,  Vx  +  1  +  Va;  —  2  =  V2  x  —  5. 

Squaring,  x  +  1  +  2\/x"^  —  x  —  2  +  x  —  2  =  2x  —  5. 

Simplifying,  Vx^  —  x  —  2  =  —  2. 

Squaring,  x^  —  x  —  2  =  4, 

Solving,  X  =  —  2  or  3. 

Verification.  —  Substituting  —  2  for  x  in  the  given  equation, 
V^^+  \/34-  V^^  =  0; 
that  is,  VZrT  +  2\/^n"- 3V^n"  =  0. 

Therefore,  —  2  is  a  root  of  the  given  equation. 
Substituting  3  for  x  in  the  given  equation, 

Vi  +  Vi  -  Vl  =  0, 

which  is  not  true  according  to  the  convention  adopted  in  the  discussion  in 
§  284.     Hence,  3  is  not  to  be  regarded  as  a  root  of  the  given  equation. 


QUADRATIC    EQUATIONS  293 

Note.  —  In  the  previous  verification,  when  only  positive  square  roots  are 
taken,  the  second  value  obtained  for  x  does  not  satisfy  the  given  equation, 
yet  this  indicates  no  error  in  the  process  of  rationalizing,  for  the  equation 
can  be  verified  by  admitting  negative  square  roots.  But,  as  explained 
in  §  283,  it  is  more  convenient  to  regard  3  as  the  root  of  the  equation 
Vx  +  1  —  Vx  —  2  —  y/2x  —  b  =  0,  which  has  for  its  first  member  one  of  the 
rationalizing  factors  of  the  given  equation. 

Solve  and  verify : 

51.  8V^  — 8aj  =  f.  53.    a;  —  1  -I-  Va;  +  5  =  0. 

52.  3a;+V^  =  5V4^-  54.    cc  —  5—  Va;  —  3  =  0. 


55.    V4  a;  +  17  +  Va;  +  1  —4  =  0. 


56.  Va;  —  1  +  V2  a;  —  1  —  Vo  «  =  0. 

57.  V2a;-7-V2^+Va;-7=0. 


58.    Va;  +  3+V4a;  +  l-Vl0a;  +  4  =  0. 


59.    V6~+x  +  Va;  -  VlO  — 4a;  =  0. 


60.  V4  aj  -  3  -  V2  a;  -I-  2  =  Va;  -  6. 

61.  V2  a;  -f  3  -  Va;  +  1  =  V5  a;  -  14. 


62.    V3a;-5+Va;  — 9=V4a;-4. 


63.    Va;  +  a^  -  Va;  -  2  a^  =  V2a;  -  5  aK 

Problems 

297.    1 .    The  sum  of  two  numbers  is  8,  and  their  product  is  15. 
Find  the  numbers. 

Solution 

Let  X  =  one  number. 

Then,  8  —  x  =  the  other. 

Since  their  product  is  15,        8  x  —  x^  =  15. 

Solving,  X  =  3  or  5, 

and  8  —  X  =  5  or  3. 

Therefore,  the  numbers  are  3  and  5. 


294 


QUADRATIC  EQUATIONS 


2.  A  party  hired  a  coach  for  $12.  In  consequence  of  the 
faihire  of  three  of  them  to  pay,  each  of  the  others  had  to  pay 
20  cents  more.     How  many  persons  were  in  the  party  ? 


Let 
Then, 


and 

Therefore, 


X 

x-3 

12 
x-3 


Solution 
the  number  of  persons, 
the  number  who  paid, 

the  number  of  dollars  each  paid, 


12 

—  =  the  number  of  dollars  each  should  have  paid. 

X 

12        1  ^  12 
x  —  3     5      X 

X  =  15  or  —  12. 


Solving, 

The  second  value  of  x  is  evidently  inadmissible.  Hence,  the  number  of 
persons  in  the  party  was  16. 

3.  A  cistern  can  be  filled  by  two  pipes  in  24  minutes.  If  it 
takes  the  smaller  pipe  20  minutes  longer  to  fill  the  cistern  than 
the  larger  pipe,  in  what  time  can  the  cistern  be  filled  by  each 
pipe  ? 


Let 
Then, 

Since 

and 

and 

Solving, 


SOLCTION 

X  =  the  number  of  minutes  required  by  the  larger  pipe. 
X  +  20  =  the  number  of  minutes  required  by  the  smaller. 

-  =  the  part  which  the  larger  pipe  fills  in  one  minute, 
=  the  part  which  the  smaller  pipe  fills  in  one  minute, 


X4-20 

^^  =  the  part  which  both  pipes  fill  in  one  minute, 

1 ^J_ 

24* 


1  + 

X     x  +  20 


X  =  40  or  -  12. 


Hence,  the  larger  pipe  can  fill  the  cistern  in  40  minutes,  and  the  smaller 
in  60  minutes. 

4.  Divide  20  into  two  parts  whose  product  is  96. 

5.  Divide  14  into  two  parts  whose  product  is  45. 

6.  A  man  purchased  a  flock  of  sheep  for  $75.  If  he  had 
paid  the  same  sum  for  a  flock  containing  3  more,  they  would 
have  cost  $  1.25  less  per  head.     How  many  did  he  purchase  ? 


QUADRATIC  EQUATIONS  295 

7.  A  rectangular  garden  is  12  rods  longer  than  it  is  wide, 
and  contains  1  acre.     What  are  its  dimensions  ? 

8.  If  each  side  of  a  square  field  were  lengthened  4  rods,  the 
area  would  be  increased  136  square  rods  more  than  \  of  it.  What 
are  the  dimensions  of  the  field  ? 

9.  A  rectangular  lot  is  8  rods  longer  than  it  is  wide.  What 
are  its  dimensions,  if  it  contains  1\  acres  ? 

10.  In  a  column  of  600  soldiers  each  file  contained  3  men  more 
than  9  times  as  many  men  as  each  rank.  How  broad  and  how 
deep  was  the  column  ? 

11.  A  party  had  a  dinner  that  cost  $60.  If  there  had  been 
5  persons  more,  the  share  of  each  would  have  been  $  1  less. 
How  many  persons  were  there  in  the  party  ? 

12.  A  man  worked  a  certain  number  of  days  for  $30.  If  he 
had  received  $1  per  day  less  than  he  did,  he  would  have  been 
obliged  to  work  5  days  longer  to  earn  the  same  sum.  How  many 
days  did  he  work  ? 

13.  Find  two  consecutive  numbers  the  sum  of  whose  squares 
is  61. 

14.  Find  two  consecutive  numbers  the  sum  of  whose  recipro- 
cals is  1^^. 

15.  A  picture  that  was  8  inches  by  12  inches  was  placed  in  a 
frame  of  uniform  width.  If  the  area  of  the  frame  was  equal  to 
that  of  the  picture,  what  was  the  width  of  the  frame  ? 

16.  A  merchant  purchased  a  quantity  of  flour  for  $  100,  and 
retailed  it  at  a  gain  of  $  1  per  barrel.  After  he  had  sold  $  100 
worth  of  it,  he  had  5  barrels  of  it  left.  How  many  barrels  did 
he  buy,  and  at  what  price  ? 

17.  A  merchant  sold  a  hunting  coat  for  $  11,  and  gained  a 
per  cent  equal  to  the  number  of  dollars  the  coat  cost  him.  What 
was  his  per  cent  of  gain  ? 

18.  A  railway  train  traveled  5  miles  an  hour  slower  than 
usual  and  was  one  hour  late  in  making  a  run  of  280  miles.  How 
many  miles  per  hour  did  it  travel  ? 


296  QUADRATIC  EQUATIONS 

19.  A  rectangular  park  56  rods  long  and  16  rods  wide  was 
surrounded  by  a  street  of  uniform  width  containing  4  acres. 
What  was  the  width  of  the  street  ? 

"20.  A  boatman  rowed  8  miles  up  a  stream  and  back  in  3  hours. 
If  the  velocity  of  the  current  was  2  miles  an  hour,  what  was  his 
rate  of  rowing  in  still  water? 

21.  A  man  who  owned  a  lot  56  rods  long  and  28  rods  wide 
constructed  a  road  around  it,  thereb}^  decreasing  the  area  of  the 
lot  2  acres.     What  was  the  width  of  the  road  ? 

22.  A  man  bought  two  lots  of  cloth  and  paid  96  shillings  for 
each.  There  were  20  yards  in  all,  and  the  number  of  shillings 
per  yard  paid  for  each  was  the  same  as  the  number  of  yards  of 
the  other.     How  many  yards  of  each  did  he  buy  ? 

23.  Find  the  price  of  eggs,  when  2  less  for  30  cents  raises  the 
price  2  cents  per  dozen. 

24.  A  and  B  started  at  the  same  time  and  traveled  toward  a 
place  75  miles  distant.  A  traveled  one  mile  an  hour  faster  than 
B  and  reached  the  place  2^  hours  before  B.  At  what  rate  did 
each  travel  ? 

25.  A  person  drew  a  quantity  of  wine  from  a  cask  filled  with 
81  gallons  of  pure  wine,  and  replaced  it  with  water.  He  then 
drew  from  the  mixture  as  many  gallons  as  he  drew  before  of  pure 
wine,  when  it  was  found  that  the  cask  contained  only  64  gallons 
of  pure  wine.     How  many  gallons  did  he  draw  each  time  ? 

26.  The  circumference  of  the  fore  wheel  of  a  coach  is  5  feet 
less  than  that  of  the  hind  wheel.  If  the  fore  wheel  makes  150 
more  revolutions  than  the  hind  wheel  in  going  a  mile,  what  is 
the  circumference  of  each  wheel  ? 

27.  Two  pipes  running  together  can  fill  a  cistern  in  24  hours. 
The  larger  pipe  can  fill  the  cistern  in  2  hours  less  time  than  the 
smaller.  How  many  hours  will  it  take  each  pipe  alone  to  fill 
the  cistern  ? 

28.  It  took  a  number  of  men  as  many  days  to  dig  a  ditch  as 
there  were  men.  If  there  had  been  6  more  men,  they  would  have 
done  the  work  in  8  days.     How  many  men  were  there  ? 


QUADRATIC  EQUATIONS  297 

EQUATIONS   IN   THE   QUADRATIC   FORM 

298.  An  equation  that  contains  but  two  powers  of  an  unknown 
number  or  expression,  the  exponent  of  one  power  being  twice 
that  of  the  other,  is  in  the  Quadratic  Form. 

Equations  in  the  quadratic  form  can  be  reduced  to  the  general 
form  aa^"  +  6a;"  +  c  =  0,  in  which  n  represents  any  number. 

Examples 

1.  Given  a;*  +  6  ar'  —  40  =  0,  to  find  the  values  of  x. 

Solution 
5C*  +  6  x2  -  40  =  0. 
Factoring,  (x^  -  4)  (a;2  +  lO)  =  0. 

.-.  x^-4  =  0  or  x2  +  10  =  0, 

and  x  =  ±2  or    ±  V—  10. 

2.  Given  x^  —  x*  =  6,  to  find  the  values  of  x. 

First  Solution 
x^  -x^  =  6. 
Completing  the  square,   x*  —  x'  +  (i^)^  =  ^^. 


Extracting  the  square 

root,         x^  -  ^  =  ±  f . 

.-.  xi  =  3  or  -  2. 

Raising  to  the  fourth  j 

power,                X  =  81    or  16. 

Second  Solution 

x2  -  xi  =  6. 

Let  X*  =  p,  then, 

x^  -  i)2,  and  p"^-  p  =  6. 

.-.  p2_p_6  =  0. 

Factoring, 

(P-3)(.i>  +  2)  =  0. 

.:  p  =  S  or  -  2 ; 

that  is, 

JF*  =  3  or  -  2. 

Whence, 

«  =  81  or  16. 

298  QUADRATIC   EQUATIONS 

3.  Solve  the  equation  a;  —  4  o;^  +  3  x^  =  0. 

Solution 

Let  x^  —  p,  then,  x»  =  p"^,  and  x  =  p\ 

Then,  p^  -ip^  +  Zp=iQ. 

Factoring,  p(p2  _  4p  +  3)  =  o. 

Whence,  J?  =  0, 

or  p2_4p  +  3_0. 

Factoring,  (p  -  1)  (p  -  3)  =  0. 

Whence,  J>  =  1  or  p  =  3. 

That  is,  x^  =  0,  1,  or  3. 

.-.  x  =  0,  1,  or  27. 

4.  Given  a:^  —  7  a;  4-  Va^  —  7  a;  +  18  =  24,  to  find  the  value  of  x. 

Solution 


x2  _  7  X  +  Vx2  -  7  X  +  18  =  24.  (1) 


Adding  18,       x2  -  7  x  +  18  +  Vx2  -  7  x  +  18  =  42.  (2) 

Put  p  for  (x2  -  7  X  +  18)2  and  p^  for  (x2  -  7  x  +  18).  (3) 

Then,  p^+p-^2-0.  (4) 

Solving,  p  =  6  or  -  7.  (6) 

That  is,  Vx2  -  7  X  +  18  =  6,  (6) 


or  Vx2  _  7  X  +  18  =  -  7.                                         (7) 

Squaring  (6),  ^2  -  7  x  +  18  =  36. 

Solving,  -,    '                   X  =  9  or  -  2. 

Squaring  (7),  x2  -  7  x  +  18  =  49. 

Solving,  X  =  1  ±  |\/T73. 

Hence,  x  =  9,   -  2,  or  J  ±  JvTTS. 

5.   Solve  the  equation  a^  —  9a^  +  8  =  0. 

Solution 

x6- 9x3 +  8  =  0.  (1) 

Factoring,  (x^  -  l)(x3  -  8)=  0.  (2) 

Therefore,  x^  -  1  =  0,  (8) 

or  358  _  8  =  0.  (4) 


QUADRATIC  EQUATIONS  299 

If  the  values  of  x  ate  found  by  ti'ansposing  the  known  terms  in  (3)  and 
(4)  and  extracting  the  cube  root  of  each  member,  only  one  value  of  x  will 
be  obtained  from  each  equation.  But  if  the  equations  are  factored,  three 
values  of  x  are  obtained. 

Factoring,  (x  -  1)  (x2  +  x  +  1)  =  0,  (5) 

and  (x-2)(x2  +  2x  +  4)  =  0.  (6) 

Writing  each  factor  equal  to  zero,  and  solving : 

From  Eq.  (5),    x  =  1.   -  i  +  ^V^S,  -  ^  -  ^V^^.  (7) 

From  Eq.  (6),    x  =  2,   -  1+  V^   -  1  -  V^^.  (8) 

Since  the  values  of  x  in  (7)  are  obtained  by  factoring  x*  —  1  =  0,  they 
may  be  regarded  as  the  three  cube  roots  of  the  number  1.  Also,  the  values 
of  X  in  (8)  may  be  regarded  as  the  three  cube  roots  of  the  number  8  (§  225). 

6.    Solve  the  equation  a;*  +  4a^— 8a;  +  3  =  0. 

Solution 

Extracting  the  square  root  of  the  first  member  as  far  as  possible, 

x*  +  4x8-8x  +  3  I  X-  +  2  X  -  2 

X* 


2x2 +  2x  1 

4x3 

4  x3  +  4  x2 

2x2 +  4X' 

-2|  -4x2-8x  +  3 
-4x2-8x  +  4 

- 1 

Since  the  first  member  lacks  1  of  being  a  perfect  square,  the  square  may 
be  completed  by  adding  1  to  each  member,  giving  the  following  equation  : 

X*  +  4  x8  -  8  X  +  4  =  1. 
Extracting  the  square  root,       x2  +  2x  —  2=±1. 

.-.  x2  +  2  X  -  3  =  0,  and  x2  +  2  X  -  1  =  0. 
Solving,  X  =  1,  -  3,   -  1  ±  V2. 

Solve  the  following  equations  : 

7.  a^-13a^  +  36  =  0.  11.  5x* +  Qa?-ll  =  0. 

8.  ic* -  25  a^  +  144  =  0.  12.  2.'c*  -  8  ar'- 90  =  0. 

9.  a;*-18a^  +  32  =  0.  13.  a;*  -  5^^  +  6=  0. 
10.    3a;*  +  5ar'-8  =  0.  14.  sc^ -|- 3 a;^ -  28  =  0. 


300  QUADRATIC   EQUATIONS 

15.  x^-Sx^  =  -2.  25.  x-^x^  +  2x^  =  0. 

16.  xi  —  x^  =  Q.  26.  5x  =  xVx  +  QVx. 

17.  a;  +  2V^  =  3.  27.  3  a;  =  rc^/^ -f- 2^^. 

18.  a;* -2x^  =  3.  28.  a;"^  -  3  -  4  x^  =  0. 

19.  ar»  +  8a;^-9  =  0.  29.  a;"^- 6x5  =  1. 

20.  xi-\-xi-2  =  0.  30.  x-^-^x^  =  2x-^. 

21.  a/x  +  3Vx  =  30.  31.  x  +  2xi  =  '6x^. 

22.  aa^»  +  6a;"  +  c  =  0.  32.  2  x  +  Vx  =  15  a; V^. 

23.  a;^-4aj-5a;^  =  0.  33.  Vx  +  5  +  6x"^  =  0. 

24.  a;^  —  ar^  —  12  a;^  =  0.  34.  x*  =  8  a;  +  7  a^ Vx. 

35.  (x-3)2  +  2(a;-3)  =  3. 

36.  (ar^  +  l)2  +  4(a^  +  l)  =  45. 

37.  (a^-4)2-3(a^-4)  =  10. 

38.  {a?-2xy-2{x'-2x)  =  S. 

39.  (x2  -  x)2  -  (x^  -x)-  132  =  0. 

40.  X—5  +  2Va;  — 5  =  8. 

41.  ^2  -  3  x  +  6  +2  Va^  -  3  a;  +  6  =  24. 

42.  x'-bx^  2y/x? -  5 a; -  2  =  10. 

43.  x2  —  X  -  Va^  — a;  +  4  —  8=0. 

44.  ar'  —  5  x  +  5 Var'  —  5  x  +  1  =  49. 

45.  a;+10  =  2Va;  +  10-}-5.  49.  a;  +  2 Vx  +  3  =  21. 

46.  a;  -  3  =  21  -  4  Vx  -  3.  50.  2  a;  -  3  V2  a;  +  5  =  -  5. 

47.  2a;-6V2x-l  =  8.  51.  ar^  +  xVx  -  72  =  0. 

48.  x  =  11  —  3  Vx  +  7.  52.  x~^  —  5  a;"^  4-  4  =  0. 


53.       /'12_;^y_^8/12_j\  33^ 


QUADRATIC  EQUATIONS  301 

=  8. 


■V- 


66.    /'l^Y-4^1--' 


58.  (x-a)^-3a^(x-a)^  +  2a^  =  0. 

59.  Find  the  three  cube  roots  of  —  1. 

60.  Find  the  three  cube  roots  of  —  8. 

61.  Find  the  four  fourth  roots  of  1. 

Solve  the  following  equations : 

62.  ic«  -  28  a^  +  27  =  0.  65.    a;*  +  2  ic^  _  g.  ^  30. 

c  66.    X*  — 4a^  +  8x  =  -3. 

63.  2:^-^  =  7. 

a^  67.   ic*  -  2  ar' +  aj  =  132. 

64.  a^-16  =  0.  68.    a.-^  -  6 a;^ _|_  27 a;  =  10. 

69.  x*  +  2a^-\-5x'  +  4:X-60  =  0. 

70.  a^  +  6a^  +  7a;2_6a,_8  =  0. 

71.  a;*- 6 ar'H- 15 a,-2-18a;  +  8  =  0. 

72.  a;*-10«8  +  35a^-50a;  +  24  =  0. 

73.  16  x*  -  8  a^  -  31  ar^  + 8  a; +15  =  0. 

74.  4a;*-4a:3-7a^  +  4a;  +  3  =  0. 
a^        a;  +  l       7 


75. 


a;  + 1         a^         12 


Suggestion.  —  Since  the  second  term  is  the  reciprocal  of  the  first,  put  p 

for  the  first  term  and  —  for  the  second. 
P 

Then,  p-l  =  i 

76.   ^+2      ^2.  78.    §±l  +  2i^=f. 

2a^  +  a;  ar^  +  4aj  +  2          5 

77     a^  +  1  I       4      ^5  0,24.1      4(a;-l)^21 

■       4      "^ar*-;-!     2*  '    a;-l        o^  _,_  ^        5 


302  QUADRATIC   EQUATIONS 

80.  _^_  +  ^:^  =  -l^.  83.    x'-2x-^- ^ =  11. 

ar  —  1         X  6  ar  —  2a;  —  6 

81.  x'  +  x  +  l--—^ -  =  %    84.    a^-a;  +  -— -^ =  7. 

ar  +  a;  +  l      d  ar  — a;  — 4 

82.  ;>?-3x+  I  =1.      85.    a;^-2a;+  ^  =4. 

ar  — 3a;  +  2  ar  — 2x  +  l 

86. 1—-^- g 3  =  0. 


SIMULTANEOUS  EQUATIONS  INVOLVING 
QUADRATICS 

299.  An  equation  whose  terms  are  homogeneous  with  respect 
to  the  unknown  numbers  is  called  a  Homogeneous  Equation. 

x2  —  a;y  +  y2  —  6^  jg3  ^  y3  —  12,  and  ax  +  ?/  =  10  are  homogeneous  equations. 

300.  An  equation  that  is  not  affected  by  interchanging  the 
unknown  numbers  involved  in  it  is  called  a  Symmetrical  Equation. 

2x^  +  xy  +  2y^  =  32,  x^  +  y^  =  28  are  symmetrical  equations. 

301.  Many  simultaneous  equations  involving  quadratics  may 
be  solved  by  the  rules  for  quadratics,  if  they  belong  to  one  of  the 
following  classes : 

1.  When  one  is  simple  and  the  other  quadratic. 

2.  When  the  unknown  numbers  are  involved  in  a  similar  manner 
in  each  equation. 

3.  WJien  each  equation  is  homogeneous  and  qxiadratic. 


I.    Simple  and  quadratic. 
1.    Solve  the  equations 


x-\-y  =  l, 
3a^  +  2/2^43^ 


Solution 

x+y  =1.  (1) 

3  «2  + 8/2  =  43.  (2) 

From  (1),  y  =  7  -X.  (3) 

Substituting  in  (2),  3a;2  +  (7  -  x)2  =  43.  (4) 


X  =  3  or  J. 

(6) 

y  =  i. 

(6) 

y  =  ¥- 

<7) 

(  when  X  =  3,  j/  =  4, 

QUADRATIC  EQUATIONS  303 

Solving, 

Substituting  3  for  x  in  (3), 

Substituting  ^  for  x  in  (3), 

That  is,  X  and  y  each  have  two  values 

[  when  X  =  ^,  ?/  =  ^^• 

Equations  of  this  class  may  be  solved  by  finding  the  value  of 
one  unknown  number  in  terms  of  the  other  in  one  equation,  and 
then  substituting  it  in  the  other. 

Solve  the  following  equations : 

V  +  2/2_20,  (x  =  6-y, 


^'    '  x  =  2y.  "    [x'  +  f  =  72. 

10x  +  y  =  3xy,  (xy{x-2y)  =  10, 

y  —  x  =  2.  1  an/  =  10. 

x'  +  xy  =  12,  ^     (3x(y  +  l)  =  12, 

x  —  y  =  2.  \Sx  =  2y. 

II.    Unknown  numbers  similarly  involved. 

{ic  4-  w  =  7 
xy  =  10. 

Solution 

x  +  y  =  7.  (1) 

xy  =  10.  (2) 

Squaring  (1),  z^  +  2xy-{-y^  =  49.  (3) 

Multiplying  (2)  by  4,  4  xy  =  40.  (4) 

Subtracting  (4)  from  (3),  x^-2xy  +  y^  =  9.  (5) 

Extracting  the  square  root,  x  —  y  =  ±  S.  (6) 

From  (1)  +  (6),  x  =  5  or  2. 

From  (1)  -  (6),  y  =  2  or  6. 

Such  equations  may  be  solved  by  the  method  illustrated  in 
example  1,  but  the  method  given  above  of  finding  the  value  of 


304  QUADRATIC   EQUATIONS 

X  —  y,  so  that  it  may  be  combined  with  the  value  of  x-{-y  to  dis- 
cover the  vahies  of  x  and  y,  is  preferable. 

f  x^  4-  y^  =  25 
9.   Solve  the  equations  \  ' 

[x  +  y  =  7. 

Solution 

x2  +  2/2  =  25. 
x  +  y  =  7. 
Squaring  (2),  x^  +  2xy  +  y^  =  49. 

Subtracting  (1)  from  (3),  2xy  =  24. 

Subtracting  (4)  from  (1),       x^  —  2  xy  +  y^  =  1. 
Extracting  the  square  root,  x  —  y  =  ±1. 

From  (2) +  (6),  x  =  4  or  .3. 

From  (2) -(6),  .  2/ =  3  or  4. 


10.    Solve  the  equations 


a!*  +  2/^  =  97, 
x  +  y  =  l. 

Solution 

x*-\-y*  =  97. 

x  +  y  =  l. 
4th  power  of  (2), 

x^  +  ^afiy  +  6  x2j/2  4.  4  xyS  +  y*  =  1. 

Subtracting  (1 ) ,  4  x^y  +  6  xV  +  4  ajys  :=  _  gg. 

Dividing  by  2,  2  x^y  +  3  x^y^  +  2  xyS  =  _  48. 

2  xy  X  square  of  (2),   2  x^  +  4  x^y'^  +  2xy^  =  2  xy. 

Subtracting  (5)  from  (6),  xV  -2xy  =  48. 

Solving  for  xy,  xy  =  —  6  or  8. 

Equations  (2)  and  (8)  give  two  pairs  of  simultaneous  equations, 

x+y=l       ^    fx+y=l 
and 


xy  =  —  6  [  xy  =  8 

Solving  these  by  previous  methods, 


x  =  3,    or    -2,    or   ^  +  |V-31,    or   ^  _  i-sA-Si. 
y=-2,   or  3,   or  ^  -  ^V-Sl,   or  J  +  ^V-  31. 


QUADRATIC  EQUATIONS  305 

Solve  the  following  equations : 

r^  +  f  =  m,  j^   rx^  +  t/^  =  i3, 

"•    \xy  =  7.  '  \x  +  y  +  xy  =  ll. 

P  +  ^  =  8,  15.  P^^^'  =  ''' 

lx  +  y  =  9,  ^,    |a^  +  .^,^  +  ,-21, 

.a^  +  2/3^243.  ■  \x'  +  xy  +  f  =  7- 


13 


III.   Homogeneous  quadratics. 

(x^-xy  +  y'  =  21, 
17.    Solve  the  equations  |^2_2a^  =  _15. 


SOLCTIOK 

x^-xy  +  y^  =  21. 

y^-2xy  =-  16. 

Assume 

x  =  vy. 

Substituting  in  (1), 

«2j/2  _  vy2  +  y-2  =  21. 

Substituting  in  (2), 

y^  —  2vy^  =  —  15. 

21 

Solving  (4)  for  y% 

^       v'^-V  +  l 

„2_      15     , 

Solving  (5)  for  y^, 

^  -2t;-l 

15                21 

Comparing  the  values  oiy,         2v-i     v^-v  +  1 

Clearing,  etc., 

5 1)2  -  19  w  +  12  =  0. 

Factoring, 

(«-3)(5v-4)  =  0. 

.-.  t)  =  3  or  |. 

Substituting  3  for  v  in  (7)  or  in  (6) ,     y  =  ±y/S   } 
and  since  x  =  vy,  *  =  ±3v3 

Substituting  |  for  v  in  (7)  or  in  (6),     y  =  ±  5 
and  since  x  =  vy,  a;  =  ±  4 

ra;  =  +3\/3,  or  -3v/3,  or  +4,  or  -4. 

Hence,  |  y  =  + VS,  or  -  V3,  or  +  6,  or  -  6 

ALG.— 20 


(1) 

(2) 
(3) 
(4) 
(5) 

(6) 
(7) 

(8) 

(9) 
(10) 

(11) 


306  QUADRATIC  EQUATIONS 

Solve  the  following  equations : 

[x'-dxy^-S.  '   \u?-Sxy  +  2y'  =  ^. 

^^     {a?  +  xy  =  12,  {x^-xy  +  f  =  21, 

'   \xy  +  2y^  =  5.  '   \^+2f=:21. 

r  a;2  +  2  2/2  =  44,  f  2  0-^  -  3  a:?/ +  2/  =  100, 

i  a;?/  —  2/^  =  8.  '    \y?  —  y^z=  75. 

302.  Many  simultaneous  equations  that  belong  to  one  or  more 
of  the  preceding  classes,  and  many  that  belong  to  none  of  them, 
may  be  readily  solved  by  special  devices. 


24.    Solve  the  equations    \ 


f  x'  +  xy  =  12, 


Solution 

a;2  +  xy  =  12. 

(1) 

xy  +  ?/2  =  4. 

(2) 

Adding,                       x'^  +  2xy  +  y'^  =  16. 

(3) 

.-.  a;  +  2/  =  +  4  or  —  4. 

(4) 

Subtracting  (2)  from  (1),      x"^  -  y^  =  g. 

(5) 

Dividing  (5)  by  (4),                  x  —  y  =  +  2  or  —  2. 

(6) 

Combining  (4)  and  (6),                  a;  =  3  or  —  H  ;  y  =  1  or  —  1. 

25.    Solve  the  equations 


x^  +  a;y  +  .V*  =  48, 
x^  —  xy  +  y^  =  12. 


Solution 

x4  +  xV-  +  2/^  =  48.  (1) 

x^  -  x?/  +  2/'^  =  12.  (2) 

Dividing  (1)  by  (2),      x'^  +  xy  +  y^  =  4.  (3) 

From  (3)  -  (2),                                xy  =  -  4.  (4) 

By  adding  (4)  to  (3),  and  subtracting  (4)  from  (2),  the  values  of  (x  +  y)^ 
and  (x  —  y)^  may  be  found  and  the  solution  readily  completed. 


QUADRATIC  EQUATIONS  307 

X^-y  =26, 


26.    Solve  the  equations 

x-y  =  2. 

Solution 

x8  -  y»  =  26.  (1) 

x-y  =  2.  (2) 

Dividing  (1)  by  (2),     z^  +  xy  +  y^  ^  IS.  (3) 

Squaring  (2),  x^  -2xy  +  y^  rz4.  (4) 

Subtracting,  dividing  by  3,  xy  =  3.  (5) 

By  adding  (5)  to  (3),  the  value  of  (x  +  yy  may  be  obtained  and  the 
solution  completed  as  in  previous  examples. 


c  .       .  ■  ix*  +  y*  =  82 

27.    Solve  the  equations    i 

I    x  —  v  =  2. 


x*  +  y*  =  82, 

y 

Solution 

a;*  +  y*  =  82.  (1) 

x-y  =  2.  (2) 

Assume                                           x  =  u  +  v,  (3) 

and                                                         y  =  u  —  V.  (4) 

Substituting  these  values  in  (1), 

M*  +  4  uH  +  6  uH^  +  4  uv^  +  V* 

+  M*  -  4  M8tj  +  6  mV  _  4  Mt;3  4-  w*  =  82,  (5) 

and  in  (2),                                          2v  =  2.  (6) 

Dividing  (5)  by  2,                  m*  +  6  mV  +  ^4  =  41.  (7) 

Dividing  (6)  by  2,                                          v  =  1.  (8) 

Substituting  1  for  i;  in  (7)  and  solving,       «  =  ±  2  or  ±  V—  10.            (9) 
Hence,  from  (3)  and  (4),       x  =  3  or  —  1  or  1  ±  V—  10, 


and  y  =  lor— 3or— 1±  V—  10. 

( x^  +  y^  +  x-^y  =  14:, 
28.    Solve  the  equations    \ 

[xy  =  3. 

Solution 

x^  +  y^  +  x+y  =  U. 
xy  =  3. 
Adding  twice  the  second  equation  to  the  first, 

x2  +  2  acy  +  j/2  4.  X  +  y  =  20. 


308 


QUADRATIC  EQUATIONS 


Completing  the  square,  {x+  yY  -\- x ->r  y  +  (^)2  =  20J. 
Extracting  the  square  root,  x  +  2/  +  i=±f. 

.•.  X  +  y  =  4  or  —  6. 
The  soUition  may  be  completed  by  solving  the  equations, 


'x  +  y  =  4 
xy  =  3 


and 


x+y =-6 
x?/  =  3 


The  student  will  doubtless  discover  many  other  methods  for 
solving  simultaneous  equations.  All  the  preceding  solutions  are 
but  illustrations  of  devices  that  are  important  only  because  they 
are  often  applicable. 

Solve  the  following  simultaneous  equations : 


29. 


30. 


31. 


32. 


33. 


34. 


35. 


36. 


37. 


x  —  y  =  5. 

3^  +  2/3  =  28, 

[x  +  y  =  4:. 

n  +  x  =  y, 
1  ic2  +  2/2  =  61. 

r  l+x  =  y, 

a^  +  ?/2  =  40, 
xy  =  12. 

x(x  +  y)  =  x, 
y(x-y)=-l. 

a^-{-3xy-f  =  A3, 
x  +  2y  =  10. 

(x^  +  xy-{-f  =  19, 

(  x^ -^  3  xy  =  y""  +  23, 
[x  +  3y  =  9. 


38. 


39. 


2x^-\-xy-5y^  =  20, 
2x-3y  =  l. 

^  +  —  =  21, 

y      y 

x-y  =  2. 


40.    -I 


[3^ 


{xy 


xy  =  48, 
-/  =  12. 


41.   ■! 


{ 21?  +  xy  =  —Q, 


42. 


43. 


44. 


45.   -! 


46. 


[xy  -\-y^  =  15. 

4  ajt/  =  96  —  x?y^, 
X  +  ?/  =  6. 

r  x'-xy  =  8, 
\xy  +  f  =  12. 

a?  —  xy  =■  Q, 
:(?  +  y^  =  61. 

{'j?  +  xy  =  ll, 
{xy  —  y^  =  12. 

\2x-y  =  2, 
\2x^  +  f  =  ^. 


47. 


48. 


49. 


50. 


51. 


52. 


53. 


54. 


55. 


66. 


57. 


58. 


QUADRATIC  EQUATIONS 

I  5  0^2  +  4  «2  =  41. 


309 


0^2  +  4  y2  =  41 

2xy-f  =  12, 
Sxy  +  oo?  =  104. 

1  a^  +  /  =  106. 

'  a^  +  a:y  +  2/*  =  84, 
,  X  —  V^  +  2/  =  6. 

'  4:0?  —  2  xy  ->ry'  =  13, 
.8  0^  +  ^  =  65. 
■6ar^  +  6t/2^13  3.y^ 
.a:2_,/2^20. 

a^  + 2^^-3(0^  +  2/)=  8, 

a;  4- 2/ +  2^  =  11- 

3xy  +  2x  +  y  =  25, 

9a;_4  ?/ 

2/        a; 

0^  +  x?/  =  40, 

27  +  2  2/2  =  3  0^. 

'  0^2  ^  a;?/ =  24, 
.  xy^  -\-  X  =  56. 

a;*  +  2/*  =  82, 
x  +  y  =  4. 

ic*  -  y^  =  369, 

ar^  -  2/2  =  9. 


f  ^  +  2/'  -  78  =  a;  +  2/, 
59.    i 

I  «?/  +  a;  +  1/  =  39. 


60 


|x-2  +  2.'r2/  +  32/-  =  43, 
*'   l2a^  +  3a;2/  +  42/''  =  62. 


61. 


62. 


63. 


64. 


65. 


x^  —  xy  +  2y^  =  46, 
a?  +  xy  +  3y^  =  111. 

ar^  -  7  a;?/  +  12  2/2  =  0, 
xy-\-3y  =  2x  +  21. 

a^  -  2/3  =  37, 
a^(2/-a;)  =  -12. 

a;  +  2/  =  25, 
V^+V2/  =  7. 

a-3  + 2/3^225  2/, 
3^2  -  2/2  =  75. 

f  a^  +  aw  +  2  2/2  =  11, 

*    l2a:2_^5^2^22. 

a^  +  2/^  =  3  a:2/  +  5, 
.x*  +  y*  =  2. 


67. 


68. 


a;^  4-  2/  = ' 

x-y 

x^y  —  an/2  _  Q 


1  (a;  -  2/)  (a^  -  2/^0  =  ^• 
I  a^  +  2/  =  a;  -  2/^  +  42, 
'   U.v  =  20. 


^^     I  a;  +  y  +  2  Va;  +  2/  =  24, 
I  a;  —  2/  +  3  Va;  —  y  =  10. 

(a^  +  2/'  +  6V^+7  =  55, 


310  QUADRATIC  EQUATIONS 


x-{-  y  _   2  X       a?  —  a?y  _ 3 
73.   I      y         x  +  y     f~x^y     S' 
[  a^  =  2/2  +  16. 

a?  —  xy  =  a?  -\-  h^, 


74.    I  ,      ^ 

y  xy  —  y  =1  2  ah, 

^^     {x-2y  =  2{a  +  h), 
[xy  +  2f  =  2b(b-a). 

\x'y  +  xy^  =  2a(a^-by 

ia  —  X  ,b  +  y__  ay  +  bx 
X  y         a^  —  b^' 

a^  +  2/2  =  2(a'  +  62). 


Problems 

303.  1.  The  sum  of  two  numbers  is  12,  and  their  product  is 
32.     What  are  the  numbers  ? 

2.  The  sum  of  two  numbers  is  17,  and  the  sum  of  their 
squares  is  157.     What  are  the  numbers  ? 

3.  The  difference  of  two  numbers  is  1,  and  the  difference  of 
their  cubes  91.     What  are  the  numbers  ? 

4.  The  sum  of  two  numbers  is  82,  and  the  sum  of  their  square 
roots  is  10.     What  are  the  numbers  ? 

5.  It  takes  52  rods  of  fence  to  inclose  a  rectangular  garden 
containing  1  acre.     How  long  and  how  wide  is  the  garden  ? 

6.  The  sum  of  the  squares  of  the  terms  of  a  fraction  is  89, 
and  the  fraction  is  |^  larger  than  its  reciprocal.  What  is  the 
fraction  ? 

7.  Find  two  numbers  such  that  their  product  is  8  greater 
than  twice  their  sum  and  48  less  than  the  sum  of  their  squares. 

8.  If  63  is  subtracted  from  a  certain  number  expressed  by 
two  digits,  its  digits  will  be  transposed;  and  if  the  number  is 
multiplied  by  the  sum  of  its  digits,  the  product  will  be  729. 
What  is  the  number  ? 


QUADRATIC  EQUATIONS  311 

9.  A  man  expended  $  6  for  canvas.  Had  it  cost  4  cents  less 
per  yard,  he  would  have  received  5  yards  more.  How  many 
yards  did  he  buy,  and  at  what  price  per  yard  ? 

10.  If  the  difference  of  two  numbers  multiplied  by  the 
greater  is  160,  and  multiplied  by  the  less  is  96,  what  are  the 
numbers  ? 

11.  A  rectangular  flower  garden  containing  54  square  rods 
was  enlarged  to  twice  its  former  size  by  making  an  addition 
of  1\  rods  on  all  sides.  What  were  the  original  dimensions  of 
the  garden? 

12.  If  it  requires  200  rods  of  fence  to  inclose  a  rectangular 
field  of  15  acres,  what  are  its  dimensions  ? 

13.  A  rectangular  field  contains  20  acres.  If  its  length  were 
20  rods  less  and  its  width  8  rods  less,  its  area  would  be  8  acres 
less.     What  are  its  dimensions  ? 

14.  A  man  found  that  he  could  buy  16  more  sheep  than  cows 
for  $  100,  and  that  the  cost  of  3  cows  was  $  15  greater  than  the 
cost  of  4  times  as  many  sheep.     What  was  the  price  of  each  ? 

15.  Eight  persons  contributed  $30  to  pay  for  a  set  of  books. 
One  half  of  the  amount  was  contributed  by  women,  and  the  other 
half  by  men,  each  man  giving  $  2  more  than  each  woman.  What 
did  each  woman  and  what  did  each  man  contribute  ? 

16.  A  man  loaned  $  1000  in  two  unequal  sums  at  such  rates 
that  both  sums  yielded  the  same  annual  interest.  The  larger 
sum  at  the  higher  rate  of  interest  would  have  yielded  $36 
annually,  the  smaller  sum  at  the  lower  rate,  $16  annually. 
What  sums  did  he  invest,  and  at  what  rates  of  interest? 

17.  If  2  is  added  to  the  numerator  and  subtracted  from  the 
denominator  of  a  certain  fraction,  the  result  will  be  the  recipro- 
cal of  the  fraction;  if  3  is  subtracted  from  the  numerator  and 
added  to  the  denominator,  the  result  will  be  ^  of  the  original 
fraction.     What  is  the  fraction  ? 

18.  The  product  of  two  numbers  is  59  greater  than  their  sum, 
and  the  sum  of  their  squares  is  170.     What  are  the  numbers  ? 


312  QUADRATIC  EQUATIONS 

19.  A  purse  contained  $50.50  in  gold  and  silver  coins.  If 
there  were  fifteen  coins,  and  if  each  gold  coin  was  worth  as  many 
dollars  as  there  were  silver  coins  and  each  silver  coin  was  worth 
as  many  cents  as  there  were  gold  coins,  how  many  coins  of  each 
kind  were  there  ? 

20.  Two  men  working  together  can  complete  a  piece  of  work 
in  6f  days.  If  it  would  take  one  man  3  days  longer  than  the 
other  to  do  the  work  alone,  in  how  many  days  can  each  man  do 
the  work  alone  ? 

21.  The  fore  wheel  of  a  carriage  makes  12  revolutions  more 
than  the  hind  wheel  in  going  240  yards.  If  the  circumference 
of  each  wheel  were  one  yard  more  than  it  is,  the  fore  wheel 
would  make  8  revolutions  more  than  the  hind  wheel  in  going  240 
yards.     What  is  the  circumference  of  each  wheel  ? 

22.  A  sum  of  money  on  interest  for  1  year  at  a  certain  per 
cent  amounted  to  $11130.  If  the  rate  had  been  1%  less  and 
the  principal  $  100  more,  the  amount  would  have  been  the  same. 
Find  the  principal  and  rate. 

23.  A  number  multiplied  by  another  composed  of  the  same 
two  digits,  but  reversed,  gives  a  product  of  4032.  If  the  first 
divided  by  the  second  is  equal  to  If,  what  are  the  numbers  ? 

24.  The  town  A  is  on  a  lake  and  12  miles  from  B,  which  is 
4  miles  from  the  opposite  shore.  A  man  rows  across  the  lake 
and  walks  to  B  in  3  hours.  In  returning  he  walks  at  the  same 
rate  as  before,  but  rows  2  miles  an  hour  less  than  before.  If  it 
takes  him  5  hours  to  return,  find  his  rates  of  walking  and  rowing. 

25.  A,  B,  and  C  started  together  to  ride  a  certain  distance.  A 
and  C  rode  the  whole  distance  at  uniform  rates,  A  two  miles  an 
hour  faster  than  C.  B  rode  with  C  for  20  miles,  and  then,  by 
increasing  his  speed  two  miles  an  hour,  reached  his  destination 
40  minutes  earlier  than  C  and  20  minutes  later  than  A.  Find 
the  distance,  and  the  rate  at  which  each  traveled.  s 

26.  Find  two  numbers  such  that  their  product  is  equal  to  the 
difference  of  their  squares,  and  the  difference  of  their  cubes  is 
equal  to  the  sum  of  their  squares. 


QUADRATIC  EQUATIONS  313 

PROPERTIES   OP  QUADRATICS 

304.    Every  quadratic  equation  may  be  reduced  to  the  form 

aa^  +  bx  +  c  =  0,  (1) 

in  which  a  is  positive  and  b  and  c  are  positive  or  negative. 
Denote  tlie  roots  by  rj  and  rg.     Then,  §  294,  Ex.  13, 


ft  +  Vy-4ac^^^^_-&-V&^-4ac^  ^2) 


2a  2a 

In  the  following  discussion  of  the  nature  of  the  roots  of  a  quadratic  equa- 
tion, the  student  should  keep  in  mind  the  distinctions  between  rational  and 
irrational,  real  and  imaginary.  For  example,  2  and  Vi  are  rational,  and  real 
also  ;  \/2  and  Vs  are  irrational,  but  real ;  V—  2  and  V—  5  are  irrational, 
and  also  imaginary. 

1.    Suppose  that  &^  —  4  ac  is  positive. 


Then,  V6^  —  4  ac  is  a  positive  real  number  and  —  V6''  —  4  ac  is 
a  negative  real  number.     Hence,  the  roots  are  real  and  unequal. 

If  6^  —  4  ac  is  a  perfect  square,  the  roots  are  rational ;  otherwise 
they  are  irrational. 

2.   Suppose  that  6^  —  4  ac  =  0. 


Then,  Vb^  —  4  etc  =  0  and  the  roots  are  real  and  equal. 

3.    Suppose  that  b^  —  4ac  is  negative. 

Then,  Vft^  —  4ac  and  —  V6^  —  Aac  are  imaginary,  and  conse- 
quently both  roots  are  imaginary. 

Principles.  —  1.  In  any  quadratic  equation  ax^  -\-bx  +  c  =  0,  if 
&^  —  4  ac  is  positive,  the  roots  are  real  and  unequal;  if  b^—4  ac=0, 
the  roots  are  real  and  equal ;  if  b^  —  4  ac  is  negative,  both  roots  are 
imaginary. 

2.  If  W  —  A:ac  is  a  perfect  square  or  is  equal  to  zero,  the  roots  are 
rational;  othenvise  they  are  irrational. 

When  the  roots  are  real,  their  signs  are  found  by  comparing  the 
values  of  b  and  c. 

If  c  is  positive,  —  &  is  numerically  greater  than  ±  V6^  —  4  ac, 
whence  both  roots  have  the  sign  of  —  6  ;  if  c  is  negative,  —  &  is 
numerically  less  than  ±  V&^  —  4  ac,  whence  r^  is  positive  and  r^  is 


314  QUADRATIC   EQUATIONS 

negative.     The  root  haviug  the  sign  opposite  to  that  of  h  is  the 
greater  numerically.     Hence, 

Principle  3.  —  If  c  is  positive,  both  roots  have  the  sign  opposite 
to  that  of  b;  if  c  is  negative,  the  roots  have  opposite  signs,  and  the 
numerically  greater  root  has  the  sign  opposite  to  that  of  b. 

The  following  are  special  cases: 

1.  If  c  =  0,  (1)  reduces  to  the  form  ax^  +  bx  =  0,  which  has  two  roots,  —  - 
and  0.  " 

2.  If  6  =  0,  (1)  becomes  the  pure  quadratic  equation  ax^  +  c  =  0,  whose 
roots  are  numerically  equal  with  opposite  signs. 

3.  If  c  =  0  and  b  =  0,  (1)  becomes  ax'^  =  0,  which  has  two  zero  roots. 

4.  If  a  =  0  or  if  a  =  0  and  6  =  0,  (1)  ceases  to  be  a  quadratic  equation. 
But  if  these  coefficients  differ  from  zero,  however  little,  (1)  is  still  a  quad- 
ratic equation  and  has  two  roots.  To  discover  the  nature  of  the  roots  in 
these  cases,  rationalize  the  numerators  in  (2). 

Then,         n  = ^/^  and  rg- ^^'  (3) 

-b  -  V62  -  4  ac  -b+  Vb^  -  4  ac 

Suppose  that  a  is  very  small  as  compared  with  b  and  c. 
Then,  the  denominator  of  ri  is  very  nearly  equal  to  —  6  —  VP,  or  to  —  2  b, 
and  the  denominator  of  ^2  is  very  small. 

Hence,  the  smaller  a   is   the  less  will  the  first  root  differ  from  —  - ,  the 

b 
root  of  the  simple  equation  bx  +  c  =  0,  and  the  greater  will  be  the  numerical 
value  of  the  second  root. 

Suppose  that  a  and  b  are  very  small  as  compared  with  c. 

Then,  both  denominators  in  (3)  are  very  small  as  compared  with  the 
numerators. 

Hence,  the  smaller  a  and  b  are  the  greater  will  both  roots  be  in  numerical 
value. 

Examples 

1.  What  is  the  nature  of  the  roots  of  a^  —  7a;  —  8  =  0? 

Solution.  —  Since  6^  _  4  ^^^  =  49  +  32  =  81  =  9^,  the  roots  are  real  and 
unequal  (Prin.  1),  and  rational  (Prin,  2).  Since  c  is  negative,  the  roots  have 
opposite  signs  and,  b  being  negative,  the  positive  root  is  the  greater  numeri- 
cally (Prin.  3). 

2.  What  is  the  nature  of  the  roots  of  Sx^  +  Tyx  +  S^O? 

Solution.  —  Since  b^  —  iac  =  25  —  36=  —  11,  both  roots  are  imaginary 
(Prin.  1). 


QUADRATIC  EQUATIONS  315 

Find  the  nature  of  the  roots  of  the  following  equations : 


3. 

ar^  —  5  a;  —  75  =  0. 

9. 

x^  +  x-2  =  0. 

4. 

X-  -\-bx-\-Q  =  0. 

10. 

4ar-4a;+i  =  0. 

5. 

a;^  +  7  a;  -  30  =  0. 

11. 

4a^-f6a;-4  =  0. 

6. 

a;-  —  3  a;  +  5  =  0. 

12. 

2  0^-9x4-4  =  0. 

7. 

a^  +  3a;-5  =  0. 

13. 

4a^4-16a;-f7  =  0. 

8. 

ar  +  X  +  2  =  0. 

14. 

9a^-M2a;  +  4  =  0. 

305.   Formation  of  quadratic  equations. 

Any  quadratic  equation,  as  aa?  +  hx  +  c  =  0,  may  be  reduced, 
by  dividing  both  members  by  the  coefficient  of  ar,  to  the  form 
7?  -\-  px  -\-  q  =  Q,  whose  roots  are 


-P+^P'-Al  and  r,^-P-^P'-AR. 
2  2 

Adding  the  roots,     i\  +  rg  =  — — -  =  —p- 

Li 

Multiplying  the  roots,  r^r^  =  ^  ~  ^^   ~ — ^i  =  q.     Hence, 

4 

306.  Principle.  —  The  sum  of  the  roots  of  a  quadratic  equa- 
tion having  the  form  a?  -\- jox  -\-  q  =  0  is  equal  to  the  coefficient  of  x 
with  its  sign  changed,  and  their  product  is  equal  to  the  absolute  term. 

Substituting  —  (r^  +  r^  for p,  and  r^rg  for  gina:;^-|-j9a;-f-g'  =  0, 

x^  —  (ri  +  r.2)x  +  ri>2  =  0. 

Expanding,  of  —  r^x  —  r^  -\-  r-^r^.  =  0. 

Factoring,  (x  —  rj)  (a;  —  rj)  =  0. 

Hence,  to  form  a  quadratic  equation  when  the  roots  are  given : 

Rule.  —  Subtract  each  root  from  x  and  place  the  product  of  the 
remainders  equal  to  zero. 

Examples 

1.    Form  an  equation  whose  roots  are  —5  and  2. 

Solution,    (ac  +  .5)  (x  -  2)  =  0,  or  x^  +  3  x  -  10  =  0. 

Or,  since  the  sum  of  the  roots  with  their  signs  changed  is  +5  —  2,  or  3,  and 
the  product  of  the  roots  is  —10  (Prin.),  the  equation  is  x^  +  3  x  —  10  =  0. 


316  GENERAL  REVIEW 

Form  equations  whose  roots  are 


2. 

6,4. 

8. 

a,  —  3  a. 

14. 

3+V2,  3-V2. 

3. 

5,  -3. 

9. 

a +  2,  a -2. 

15. 

_2-V5,  -2+V5. 

4. 

3,  -i. 

10. 

6  +  1,  &-1. 

16. 

2  ±  3V|. 

5. 

if 

11. 

a  +  h,  a—  b. 

17. 

-i(3±V6). 

6. 

-2,  - 

h 

12. 

Va  —  Vft,  V&. 

18. 

i(-l±V2). 

7. 

-h- 

t- 

13. 

i(«iV&). 

19. 

a(2±2V5). 

GENERAL  REVIEW 

307.    1 .  Add     X  -\/y  +  y  Vic  +  v'a^,    x^y^  —  -y/x^y  —  Va^,   Va^ 
—  Vic?/^  —  V^,  and  y  Vic  —  ic  V4^  —  V9  a-t/.  . 

2.  From  the  sum  of  2a  +  36  —  3^/  and  2 1/  —  a  —  3 6  sub- 
tract (a  —  6  —  y)  —  (a  +  &  +  y). 

3.  What  number  must  be  added  to  a  to  give  b  —  a? 

4.  If  a  =  2  and  6  =  3,  find  the  value  of 

a  +  26         b         a?  —  o?b 

a  a—  b      2a—  b 

5.  What  number  must  be  subtracted  from  a  —  6  to  give 
b-a  +  c? 

6.  Simplify  a  — {6  — a-[a  — 6— (2a  +  6)  +  (2a-6)-a]-6}. 

7.  A  grocer  sold  m  pounds  of  sugar  at  a  cents  a  pound,  and 
a  pounds  of  tea  at  6  cents  a  pound.  If  the  sugar  cost  him  b  cents 
a  pound  and  the  tea  m  cents  a  pound,  what  was  his  gain  by  the 
transaction  ? 

8.  Multiply  a^  +  2a^b  +  2  ab^  +  6^  by  a^-2  a^b  +  2  aft^  -  6^ 

9.  Multiply  a;"-^  —  2  2/"-^  by  2a;  +  2/^. 

10.  Multiply  xVx-i-xVy  +  y^/x  +  y^y  by  y/x  —  Vy- 

a  a+b  a  a+b 

11.  Multiply  2x^-5y'^  by  2  a^^  +  5  y~^. 

12.  Expand  (of  -  y"")  (x"  +  y")  (ct^"  +  y^). 

13.  Divide  x*  —  y*hy  x  —  y. 


GENERAL  REVIEW  317 

14.  Divide  ic*  —  3  a^  —  20  by  a;  —  2,  using  detached  coefficients. 

15.  Prove  that  x^—b^  is  divisible  by  x  +  6. 

16.  Divide  {a-\-b)  +  x  by  (d  +  6)*  +  ici 

17.  Factor  9x^-12  a; +  4. 

18.  Factor  9ar'  +  9a;4-2. 

19.  Factor  a^  —  3  a;  +  2. 

20.  Prove  that  a;  —  a  is  a  factor  of  a;"  +  3  ax"'^  —  4  a". 

21.  Separate  a^^  —  1  into  six  rational  factors. 

22.  Factor  4  (od  +  bcf  -(a^-b^-(^+  cpf. 

23.  Find  the  H.  C.  D.  of  a^  -  /,  a^  +  2  xy  +  y%  and  y^  +  xy. 

24.  Find  the  H.  C.  D.  of  3a^-a;-2  and  6x2  +  a;-2. 

25.  Find  the  H.  C.  D.  of  4  x'  -  11  a:^  +  11  x  -  12, 

2  X*  4-  a:^  —  4  a:^  +  7  a;  —  15,  and  2  a;*  +  a^  —  a:  — 12. 

26.  Find  the  L.  C.  M.  of  4  a^ftx,  6  aWf,  and  2  oxy. 

27.  Find  the  L.  C.  M.  of  a:^  —  y^,  x-\-y,  and  xy  —  y^. 

28.  Reduce  — — —  to  its  lowest  terms. 

3ar^-4a;-4 

29.  Reduce  —-^ „         to  its  lowest  terms. 

x^  -  2  aj^*  +  1 

^2 yi g2 2  be 

30.  Reduce  — to  its  lowest  terms. 

a2-62  +  c2  +  2ac 

31.  Simplify -^  - -^- +     ^ 


X  +  1        1  —  X        X?  —  1 

32.    Simplify -^±|-  +  -£::i|-  + 4^. 
^     •^2a;-22/     2a;  +  22/     y^-a;* 

^^"      ^°^P  ^  ^  (a  -  6)  (6  _  c)  ~  (c  -  6)(c  -  a)  "^  (c-  a)(a-  6)* 

34.  Simplity(a  +  l)(a'  +  i)(a-l^ 

35.  Simplify = = — 

X-\--  X 

X  X 


8 

GENERAL   REVIEW 

36 

Simplify 

[  "  11     M 

1  +  1            «'  +  ! 

X 

H- 

X 

-  x~ 

1 

1-1 

I           X 

x-1 

37. 

Prove  that  ^x£  =  ^. 
h      q      hq 

38. 

Prove  tlial 

a     m      an 
b      n      bm 

39.  Divide  ^-^  by  -1— J-. 

V^    V^       A/y    Vx 

40.  Raise  a  —  &  to  the  seventh  power. 

41.  Expand  (2  a  4-  3  b)*. 

42.  Expand  ( V^  +  -^f. 

43.  Square  Va  +  6  —  c. 

44.  Extract  the  square  root  of  a^+2  aVab-\-3  ab+2  bVc^+b\ 

45.  Extract  the  cube  root  of  8  a^  -  36  a^b  +  54  ab^  -  27  b^ 

46.  Extract  the  square  root  of  a  +  &  to  four  terms. 

47.  Find  the  sixth  root  of  4826809. 

48.  Reduce  Vf  to  its  simplest  form. 

49.  Reduce  V25  a*  to  its  simplest  form. 

50.  Find  the  approximate  value  of • 

51.  Multiply  2  +  a/8  by  1  -  V2. 

52.  Simplify  ^  +  ^  A 

V6  +  2 

53.  Prove  that  oiP  =  l. 

54.  Prove  that  ax'"  =  — • 

ar 

55.  Prove  that  x^  =  i^c? ;  also  that  x^  =  (-^f. 

56.  Findthevalueof  125^;  of  (^—Vl 

'        U2; 


GENERAL  REVIEW 


319 


Solve  the  following  equations : 


59. 
60. 

61. 
62. 


68. 


69. 


70. 


71. 


72. 


73. 


74. 


57. 


58. 


+ 


+ 


a  —  b 

X 

1 

X 

b 


+ 


a-\-b 

1 

X 

c 


+ 


a  +  b 


=  0. 


mx^  —  nx  =  mn. 

X*  +  -  = 

2       2 

(1  +  xy+  (1  -  xy=  242. 
1  +  a; 


1  +a;+Vl  +  a^ 


67. 

r  a;  +  2/  =  8, 

^  +  z  =  4, 
[  z  -{-  X  =  6. 

fl      1      ..^ 

-  +  -  =  10, 
\  X     y 

\'-  +  '  =  10. 

[x      y 

{2x-\-^y  +  z  =  % 
\x  +  2y  +  Zz  =  13, 
[Sx-\-  y  +  2z  =  ll. 

(ax  +  y  +  z  =  2(a  +  l), 
I  x  +  ay  +  z  =  3a  +  l, 
[  a;  +  2/  -f  az  at  a*  -j-  3. 

■  ar^  +  a^  =  24, 
.y^  +  xy  =  12. 

r  a^  +  3  07/  =  7, 
la^  + 42/2^13^ 

a^?/  +  a^  =  6, 
a;3  +  /  =  9. 


63. 
64. 

65. 

66. 

3. 

75. 


y/x  —  9  =  V^  —  1. 


ar^  +  Va^  +  16  =  14. 

x  +  ar^  +  (1  +  a;  +  a^)2  =  55. 
1  +  a; 


76. 


77. 


78. 


79. 


80. 


81. 


82. 


83. 


84. 


1  —  a;  +  Vl  +a^ 
'  a^  +  a;  =  26  —  ?/2  -  2/, 
an/ =  8. 

|V^=12, 

{x-\-y  —  Va;  +  y  =  20. 
[a:2-2/='  =  7, 

a^  -  2/*  =  175. 

a^  -  a^2  =  -  6, 

a;  —  a^  =  9. 

xy=zx  +  y, 

ar'  +  2/2  ^  8. 

ar*?/"  —  4  a;^/  =  5, 

x-2  +  42/2  =  29. 

2a^  +  2/  =  9a!2/, 

a;  +  2/  =  3. 

a^  +  a;?/  +  /  =  189, 

X  +  Va^  -f  2/  =  21. 
■  a;^  +  2/'  =  4, 
,  iK^  +  2/  =  16. 
j  V^  -  V2/  =  f  (»  -  y), 


320  GENERAL   REVIEW 

85.  A  and  B  hired  a  carriage  for  themselves  and  four  friends. 
If  all  had  paid,  A  and  B  would  each  have  had  4  dollars  less  to 
pay.     What  was  the  cost  of  hiring  the  carriage  ? 

86.  What  number  is  that  to  which  if  12  is  added  and  from 
^  of  the  smn  12  is  subtracted,  the  remainder  is  12  ? 

87.  A  grocer  has  two  kinds  of  sirup  worth  50  and  80  cents 
per  gallon  respectively.  How  many  gallons  of  each  must  he  take 
to  make  a  mixture  of  45  gallons  worth  60  cents  a  gallon  ? 

88.  How  many  dimes  and  how  many  quarters  must  be  taken 
so  that  18  coins  are  worth  $  3  ? 

89.  In  a  certain  weight  of  gunpowder  the  saltpeter  was  5 
pounds  more  than  half  the  weight,  the  sulphur  2  pounds  less 
than  a  fifth,  and  the  charcoal  1  pound  more  than  a  tenth.  Find 
the  number  of  pounds  of  each. 

90.  How  far  down  a  river  whose  current  runs  3  miles  an  hour 
can  a  steamboat  go  and  return  in  8  hours,  if  its  rate  of  sailing  in 
still  water  is  12  miles  an  hour  ? 

91.  A  woman  being  asked  what  she  paid  for  her  eggs,  replied, 
"  Six  dozen  cost  as  many  cents  as  I  can  buy  eggs  for  32  cents." 
What  was  the  price  per  dozen  ? 

92.  A  gentleman  had  not  room  in  his  stables  for  8  of  his 
horses,  so  he  built  an  additional  stable  ^  the  size  of  the  other, 
and  then  had  room  for  8  horses  more  than  he  had.  How  many 
horses  had  he  ? 

93.  In  a  mass  of  copper,  lead,  and  tin,  the  copper  was  5 
pounds  less  than  half  the  whole  in  weight,  and  the  lead  and  tin 
each  5  pounds  more  than  \  of  the  remainder.  Find  the  weight 
of  each. 

94.  At  what  time  between  4  and  5  o'clock  do  the  hands  of 
a  clock  make  a  straight  line  ? 

95.  A  person  who  can  walk  n  miles  an  hour  has  a  hours  at  his 
disposal.  How  far  may  he  ride  in  a  coach  that  travels  m  miles 
an  hour  and  return  on  foot  within  the.  allotted  time  ? 


GENERAL   REVIEW  321 

96.  A  merchant  sold  half  a  car  load  more  than  half  his  grain ; 
then  he  sold  half  a  car  load  more  than  half  the  remainder,  when 
he  found  that  if  he  could  sell  half  a  car  load  more  than  half  of 
what  he  still  had,  he  would  have  none  left.  How  many  car 
loads  of  grain  had  he  ? 

97.  Four  years  ago  A's  age  was  \  of  B's,  and  4  years  hence 
it  will  be  I  of  B's  age.     What  is  the  age  of  each  ? 

98.  A  person  being  asked  the  time  of  day  replied  that  the 
time  past  noon  was  f  of  the  time  to  midnight.  What  was  the 
time  of  day  ? 

99.  If  3  is  added  to  each  term  of  a  certain  fraction,  the  value 
of  the  fraction  will  be  |^ ;  if  3  is  subtracted  from  each  term,  the 
value  will  be  |.     What  is  the  fraction  ? 

100.  A  boatfaian  rows  such  a  distance  down  a  stream  that  it 
takes  him  4  hours  to  return.  If  it  takes  him  2  hours  to  row 
down  and  the  current  is  2  miles  an  hour,  what  is  his  rate  of 
rowing  in  still  water  ? 

101.  A  man  received  $  2.50  per  day  for  every  day  he  worked, 
and  he  agreed  to  forfeit  $  1.50  for  every  day  he  was  idle.  If  he 
worked  3  times  as  many  days  as  he  was  idle  and  received  $  24, 
how  many  days  did  he  work  ? 

102.  A  jeweler  has  two  silver  cups,  and  a  cover  worth  $  1.50. 
The  first  cup  with  the  cover  on  it  is  worth  1^  times  as  much  as 
the  second  cup,  and  the  second  cup  with  the  cover  on  it  is  worth 
\\  as  much  as  the  first  cup.     Find  the  value  of  each  cup. 

103.  Some  smugglers  discovered  a  cave  that  would  exactly 
hold  their  cargo,  which  consisted  of  13  bales  of  cotton  and  33 
casks  of  wine.  While  they  were  unloading,  a  revenue  cutter 
hove  in  sight,  and  they  sailed  away  with  9  casks  and  5  bales, 
leaving  the  cave  two  thirds  full.  How  many  bales,  or  how  many 
casks,  would  the  cave  hold  ? 

104.  Twenty-eight  tons  of  goods  are  to  be  transported  in  carts 
and  wagons,  and  it  is  found  that  it  will  require  15  carts  and  12 
wagons,  or  else  24  carts  and  8  wagons.  How  much  can  each  cart 
and  each  wagon  carry  ? 

ALG.  —  21 


322  GENERAL  REVIEW 

105.  There  is  a  number  whose  three  digits  are  the  same.  If  7 
times  the  sum  of  the  digits  is  subtracted  from  the  number,  the 
remainder  is  180.     What  is  the  number  ? 

106.  A  and  B  can  do  a  piece  of  work  in  m  days,  B  and  C  in  w 
days,  A  and  C  in  p  days.  In  what  time  can  all  together  do  it  ? 
How  long  will  it  take  each  alone  to  do  it  ? 

107.  Two  passengers  together  have  400  pounds  of  baggage 
and  are  charged,  for  the  excess  above  the  weight  allowed  free,  40 
and  60  cents  respectively.  If  the  baggage  had  belonged  to  one 
of  them,  he  would  have  been  charged  $  1.50.  How  much  baggage 
is  one  passenger  allowed  without  charge  ? 

108.  Divide  20  into  two  parts  such  that  the  sum  of  the  two 
fractions  formed  by  dividing  each  part  by  the  other  is  4^. 

109.  It  takes  1000  square  tiles  of  a  certain  size  to  pave  a  hall, 
or  1440  square  tiles  whose  dimensions  are  one  inch  less.  Find 
the  area  of  the  hall  floor. 

110.  The  sum  of  two  numbers  is  16,  and  the  difference  of  their 
squares  is  128.     What  are  the  numbers  ? 

111.  Find  two  numbers  such  that  their  sum,  their  product,  and 
the  difference  of  their  squares  are  all  equal. 

112.  Divide  25  into  two  parts  such  that  the  difference  of  their 
square  roots  is  1. 

113.  The  difference  of  two  numbers  is  6,  and  their  product 
is  equal  to  twice  the  cube  of  the  less  number.  What  are  the 
numbers  ? 

114.  It  took  a  number  of  men  as  many  days  to  pave  a  side- 
walk as  there  were  men.  Had  there  been  3  men  more,  the  work 
would  have  been  done  in  4  days.     How  many  men  were  there  ? 

115.  The  product  of  two  numbers  is  8,  and  the  sum  of  their 
squares  is  14  greater  than  the  sum  of  the  numbers.  What  are 
the  numbers  ? 

116.  A  rectangular  lawn  50  feet  long  and  40  feet  wide  has  a 
walk  of  uniform  width  around  it.  If  the  area  of  the  walk  is  64 
square  yards,  what  is  its  width  ? 


GENERAL   REVIEW  323 

117.  A  merchant  sold  goods  for  56  dollars  and  gained  as  many 
hundredths  of  the  cost  as  there  were  dollars  in  the  cost.  Find 
the  cost  of  the  goods. 

118.  A  person  swimming  in  a  stream  that  runs  1^  miles  per 
hour  finds  that  it  takes  him  3  times  as  long  to  swim  a  certain 
distance  up  the  stream  as  it  does  to  swim  the  same  distance  down. 
What  is  his  rate  of  swimming  in  still  water  ? 

119.  A  drover  bought  some  oxen  for  $  900.  After  5  had  died, 
he  sold  the  rest  at  a  profit  of  $  20  each  and  thereby  gained  $  350. 
How  many  oxen  did  he  buy  ? 

120.  A  detachment  from  an  army  was  marching  in  regular 
column  with  5  men  more  in  depth  than  in  front.  On  approaching 
the  enemy,  the  front  was  increased  by  845  men,  and  the  whole 
was  thus  drawn  up  in  5  lines.  Find  the  number  of  men  in  the 
detachment. 

121.  A  round  iron  bar  weighed  36  pounds.  If  it  had  been 
1  foot  longer  and  of  uniform  diameter,  each  foot  of  it  would  have 
weighed  \  a  pound  less.  Find  the  length  of  the  iron  bar  and  its 
weight  per  foot. 

122.  A  farmer  has  two  cubical  granaries.  The  side  of  one  is 
3  yards  longer  than  the  side  of  the  other,  and  the  difference  of 
tlieir  solid  contents  is  117  cubic  yards.  What  is  the  length  of 
the  side  of  each  ? 

123.  Two  workmen,  A  and  B,  were  employed  at  different 
wages.  At  the  end  of  a  certain  number  of  days  A  received  $  30, 
but  B,  who  had  been  idle  two  days  in  the  meantime,  received 
only  $  19,20.  If  B  had  worked  the  whole  time,  and  A  had  been 
idle  two  days,  they  would  have  received  equal  sums.  Find  the 
number  of  days,  and  the  daily  wages  of  each. 

124.  By  traveling  5  miles  an  hour  less  than  its  usual  rate  a 
train  was  50  minutes  late  in  running  300  miles.  Find  the  usual 
rate  of  speed  and  the  time  usually  required  to  make  the  trip. 

125.  Find  two  numbers  such  that  their  sum,  their  product,  and 
the  sum  of  their  squares  are  all  equal. 


324  GENERAL  REVIEW 

126.  A  merchant  bought  two  lots  of  tea,  paying  for  both  $34. 
One  lot  was  20  pounds  more  than  the  other,  and  the  number  of 
cents  paid  per  pound  was  in  each  case  equal  to  the  number  of 
pounds  bought.     How  many  pounds  of  each  did  he  buy  ? 

127.  A  and  B  hired  a  pasture  into  which  A  put  4  horses,  and 
B  as  many  as  cost  him  18  shillings  per  week.  Afterward  B  put 
in  2  additional  horses,  and  found  that  he  must  pay  20  shillings 
per  week.     How  much  was  paid  for  the  pasture  per  week  ? 

128.  By  lowering  the  selling  price  of  apples  1  cent  a  dozen, 
an  apple  woman  finds  that  she  can  sell  60  more  than  she  used  to 
sell  for  60  cents.  At  what  price  per  dozen  did  she  sell  them  at 
first? 

129.  A  and  B  are  two  stations  300  miles  apart.  Two  trains 
start  at  the  same  time,  one  from  A,  the  other  from  B,  and  travel 
to  the  opposite  station.  If  the  first  train  reaches  B  9  hours  after 
the  trains  meet,  and  the  second  train  reaches  A  4  hours  after 
they  meet,  when  do  they  meet,  and  what  is  the  rate  of  each  train  ? 

130.  If  a  carriage  wheel  14|  feet  in  circumference  takes  one 
second  longer  to  revolve,  the  rate  of  traveling  will  be  2|  miles 
less  per  hour.     How  fast  is  the  carriage  traveling  ? 

131.  A  railway  train,  after  traveling  2  hours,  was  detained  1 
hour  by  an  accident.  It  then  proceeded  at  f  of  its  former  rate, 
and  arrived  7f  hours  behind  time.  If  the  accident  had  occurred 
50  miles  farther  on,  the  train  would  have  arrived  6^  hours  behind 
time.     What  was  the  whole  distance  traveled  by  the  train  ? 

132.  A  person  rents  a  certain  number  of  acres  of  land  for 
$  200.  He  retains  5  acres  for  his  own  use  and  sublets  the  rest 
at  $  1  an  acre  more  than  he  gave.  If  he  receives  $  10  more  than 
he  pays  for  the  whole,  how  many  acres  does  he  rent,  and  at  what 
rate  per  acre  ? 

133.  A  and  B  left  Chicago  and  walked  in  the  same  direction 
at  uniform  rates.  B  started  2  hours  after  A  and  overtook  him 
at  the  30th  milestone.  Had  each  traveled  half  a  mile  more  per 
hour,  B  would  have  overtaken  A  at  the  42d  milestone.  At  what 
rate  did  each  travel  ? 


RATIO    AND    PROPOKTION 


308.  1.  What  is  the  relation  of  10  a;  to  5  a;  ?  of  3  a;  to  12  a;  ?  of 
8a  to  2a?  ofl4mto7m?  of  4a  to  8a?  of  26  to  66? 

2.  In  finding  the  relation,  or  ratio,  of  10  a  to  5  a,  which  is  the 
dividend,  the  number  that  precedes,  or  the  number  that  follows  f 
Which  is  the  divisor  ? 

3.  What  is  the  ratio  of  a  to  6  ?  Since  6  may  not  be  exactly- 
contained  in  a,  how  may  the  ratio  be  expressed  ? 

4.  Since  the  ratio  of  two  numbers  may  be  expressed  in  the 
form  of  a  fraction,  what  operations  may  be  performed  upon  the 
terms  of  a  ratio  without  changing  the  ratio  ? 

309.  The  relation  of  two  numbers  that  is  expressed  by  the 
quotient  of  the  first  divided  by  the  second  is  called  their  Ratio. 

310.  The  Sign  of  Ratio  is  a  colon  (:). 

A  ratio  is  also  expressed  in  the  form  of  a  fraction. 

The  ratio  of  a  to  6  is  written  a  :  6  or  -• 

h 
The  colon  is  sometimes  regarded  as  derived  from  the  sign  of  division  by 
omitting  the  line. 

311.  The  first  term  of  a  ratio  is  called  the  Antecedent. 
It  corresponds  to  a  dividend,  or  numerator. 

312.  The  second  term  of  a  ratio  is  called  the  Consequent. 
It  corresponds  to  a  divisor,  or  denominator. 


313.    The  antecedent  and  consequent  form  a  Couplet. 

In  the  ratio  a  :  6,  or  -,  a 
b 

terms  a  and  b  form  a  couplet. 


In  the  ratio  a :  6,  or  -,  a  is  the  antecedent,  h  the  consequent,  and  the 
b 


325 


326  RATIO  AND  PROPORTION 

314.  The  ratio  of  the  reciprocals  of  two  numbers  is  called  the 
Reciprocal,  or  Inverse  Ratio   of  the  numbers. 

It  may  be  expressed  by  interchanging  the  terms  of  the  ratio 
of  the  numbers. 

The  inverse  ratio  of  a  to  6  is  -  :  - .     Since  --=--  =  -,  the  inverse  ratio  of 
^  ah  aba 

a  to  b  may  be  written  - ,  or  6  :  a. 
a 

315.  The  ratio  of  the  squares  of  two  numbers  is  called  the 
Duplicate  ratio ;  the  ratio  of  their  cubes,  the  Triplicate  ratio ;  the 
ratio  of  their  square  roots,  the  Subduplicate  ratio;  the  ratio  of 
their  cube  roots,  the  Subtriplicate  ratio  of  the  numbers. 

The  duplicate  ratio  of  a  to  6  is  a^  :  b^ ;  the  triplicate  ratio,  a^  :  b^;  the 
subduplicate  ratio,  Va  :  y/b  ;  the  subtriplicate  ratio,  \/a  :  y/b. 

316.  Principle.  — Multiplying  or  dividing  both  terms  of  a  ratio 
by  the  same  number  does  not  change  the  ratio. 

Examples 

1.  What  is  the  ratio  of  8  m  to  4  m  ?  of  4  m  to  8  m  ? 

2.  Express  the  ratio  of  6  :  9  in  its  lowest  terras;  12x:lQy\ 
am  :  bm ;  20  ab:  10  be ;  (m  +  w)  :  ('^n^  —  n^)- 

3.  Which  is  the  greater  ratio,  2:3  or  3:4?  4:9  or  2:5? 

4.  What  is  the  ratio  oi^to\?  |  to  ^  ?'  |  to  f  ? 
Suggestion.  —  When  fractions  have  a  common  denominator,  they  have 

the  ratio  of  their  numerators. 

5.  Reduce  a :  b  and  x:y  to  ratios  having  the  same  consequent. 

6.  When  the  antecedent  is  6  a;  and  the  ratio  is  ^-,  what  is  the 
consequent  ? 

317.  It  is  evident  from  §  316,  that  the  ratio  of  two  rational 
fractions  may  be  expressed  by  the  ratio  of  two  integers. 

Eor  example,  —  ;  _  may  be  reduced  to  the  form  —  xny:  -xny, 
,  ?i     V  n  y 

or  my :  bn.  ^  ^ 

But  the  ratio  of  two  numbers,  when  one  is  rational  and  the 
other  irrational  or  when  they  are  dissimilar  surds,  cannot  be 
expressed  by  the  ratio  of  two  integers. 

Thus,  the  ratio  V2  :  3  cannot  be  expressed  by  any  two  integers- 


RATIO  AND  PROPORTION  327 

318.  If  the  ratio  of  two  numbers  can  be  expressed  by  the  ratio 
of  two  integers,  the  numbers  are  called  Commensurable  Numbers, 
and  their  ratio  a  Commensurable  Ratio. 

319.  If  the  ratio  of  two  numbers  cannot  be  expressed  by  the 
ratio  of  two  integers,  the  numbers  are  called  Incommensurable 
Numbers,  and  their  ratio  an  Incommensurable  Ratio. 

•        r            \/2      1.414213+ 
The   ratio    v2  :  3  =  -—  =  — — cannot  be  expressed    by  any  two 

o  o 

integers,  because  there  is  no  number  that,  used  as  a  common  measure,  will 
be  contained  in  both  V2  and  3  an  integral  number  of  times.  Hence,  v^  and 
3  are  incommensurable,  and  \/2  :  3  is  an  incommensurable  ratio. 

It  is  evident  that  by  continuing  the  process  of  extracting  the  square  root 
of  2,  the  ratio  V2  : 3  may  be  expressed  by  two  integers  to  any  desired  de- 
gree of  approximation,  but  never  with  absolute  accuracy. 

320.  1.  What  two  numbers  have  the  same  relation  to  each 
other  as  2  to  3  ? 

2.  Name  several  couplets  that  express  the  same  ratio  as  2:5. 
How  may  it  be  indicated  that  the  ratio  of  2  to  5  is  the  same  as 
that  of  2  a  to  5  a  ? 

3.  What  number  has  the  same  ratio  to  12  a  that  5  6  has  to  3  6  ? 

4.  What  number  has  the  same  ratio  to  10  a  that  10  a  has  to  2  ? 
How  does  this  number  times  2  compare  with  10  a? 

321.  An  equality  of  ratios  is  called  a  Proportion. 

3  :  10  =  6  :  20  and  a  :x  =  b  :  y  are  proportions. 

The  double  colon  (: ;)  is  often  used  instead  of  the  sign  of 
equality. 

The  double  colon  has  been  supposed  to  represent  the  extremities  of  the 
lines  that  form  the  sign  of  equality. 

The  proportion  a  :  b  =  c  :  d,  ov  a :  b : :  c  :  d,  is  read,  the  ratio  of  a 
to  b  is  equal  to  the  ratio  of  c  to  d,  or  a  is  to  6  as  c  is  to  d. 

322.  The  antecedents  and  consequents  of  the  ratios  that  form 
a  proportion  are  called  the  Antecedents  and  Consequents,  respec- 
tively, of  the  proportion. 

In  a  :  b  =  r. :  d,  the  antecedents  of  the  proportion  are  a  and  c,  and  the 
consequents  are  b  and  d. 


328  RATIO  AND  PROPORTION 

323.  The  first  and  fourth  terms  of  a  proportion  axe  called  the 
Extremes  of  the  proportion. 

In  the  proportion  a  :  6  =  c  :  (Z,  the  extremes  are  a  and  d. 

324.  The  second  and  third  terms  of  a  proportion  are  called  the 
Means  of  the  proportion. 

In  the  proportion  a  -.h  =  c  -.d,  the  means  are  h  and  c. 

325.  The  terms  of  a  proportion  are  also  called  Proportionals. 
In  the  proportion  a:  b  =  b  :  c,  b  is  called  a  Mean  Proportional 
between  a  and  c,  and  c  is  called  a  Third  Proportional  to  a  and  b. 

In  the  proportion  a  :  &  =  c  :  d,  d  is  called  a  Fourth  Proportional 
to  a,  b,  and  c. 

Since  a  proportion  is  an  equality  of  ratios  each  of  which  may 
be  expressed  as  a  fraction,  a  proportion  may  be  expressed  as  an 
equation  each  member  of  which  is  a  fraction.  Hence,  it  follows 
that: 

326.  General  Principle.  —  The  changes  that  may  be  made 
upon  a  proportion  without  destroying  the  equality  of  its  ratios  are 
based  upon  the  changes  that  may  be  made  upon  the  members  of  an 
equation  without  destroying  their  equality  and  upon  the  terms  of 
a  fraction  without  altering  the  value  of  the  fraction. 

PRINCIPLES    OP   PROPORTION 

327.  1.  Let  any  four  numbers  form  a  proportion,  as  a  :  6  =  c :  d. 

2.  Express  the  proportion  as  a  fractional  equation. 

3.  If  this  equation  is  cleared  of  fractious,  what  terms  of  the 
proportion  does  the  first  member  contain  ?  the  second  member  ? 

Principle  1.  —  In  any  proportion  the  product  of  the  extremes 
is  equal  to  the  product  of  the  means. 

If  a  :  b  =  c  :  d,  then,  ad  =  fee. 

Since  a  mean  proportional  serves  as  both  means  of  a  proportion, 
if  a:b  =  b:  c,  b^  =  ac,  ov  b  =  ^/ac.     Hence, 

The  mean  proportional  between  two  numbers  is  equal  to  the  square 
root  of  their  product. 


RATIO  AND  PROPORTION  329 

Principle  1  may  be  established  as  follows : 
Let  a  -.b  =  c  :d  represent  any  proportion. 

Then,  §310,  J  =  |. 

0     a 

Clearing  of  fractions,  ad  =  he. 

Therefore,  the  product  of  the  extremes  is  equal  to  the  product  of  the 
means. 

NuMEKicAL  Illustration 

2  :  5  =  6  :  15. 

2  X  15  =  5  X  6. 

30  =  30. 

328.  1.  Transform  the  proportion  a:6  =  c:d  in  accordance 
with  Prin.  1. 

2.  Since  ad  =  he,  how  may  the  value  of  a  be  found  ?  the  value 
of  d  ?     What  terms  of  the  proportion  are  a  and  d  ? 

3.  How,  then,  may  either  extreme  of  a  proportion  be  found  ? 
How  may  either  mean  be  found  ? 

Principle  2. —JE^iY^er  extreme  of  a  proportion  is  equal  to  the 
product  of  the  means  divided  by  the  other  extreme.  Either  mean  is 
equal  to  the  product  of  the  extremes  divided  by  the  other  mean. 

If  a:b  =  c:d,  then,  a  =  — ,    b  =  —,  etc. 

d  c 

Demonstrate  Prin.  2,  and  give  numerical  illustrations. 

329.  1.  If  ad  =  be,  what  will  be  the  resulting  proportion  when 
both  members  are  divided  by  bd  and  reduced  ? 

2.  What  do  the  factors  of  ad,  the  first  member  of  the  equation, 
form  in  the  proportion  ?     What  do  the  factors  of  be  form  ? 

Principle  3.  —  If  the  produet  of  two  numbers  is  equal  to  the 
product  of  two  other  numbers,  one  pair  of  them  may  be  made  the 
extremes  and  the  other  pair  the  means  of  a  proportion. 

If  ad  =  be,  then,  a  :  b  =  c  :  d,   or  b  :  a  =  d  :  c,  etc. 

Demonstrate  Prin.  3,  and  give  numerical  illustrations. 


830  RATIO  AND  PROPORTION 

330.  1.  Change  the  proportion  a:b  =  c:d  into  an  integral 
equation  by  Prin.  1. 

2.  Divide  the  members  of  this  equation  by  cd  and  reduce. 

3.  What  terms  of  the  given  proportion  now  form  the  first 
couplet  ?   the  second  couplet  ? 

Principle  4.  —  If  four  numbers  are  in  proportion,  the  ratio 
of  the  antecedents  is  equal  to  the  ratio  of  the  consequents  ;  that  is, 
the  numbers  are  in  proportion  by  Alternation. 

If  a  :  6  =  c  :  d,  then,  a  •.c  =  h  -.d. 

Demonstrate  Prin.  4,  and  give  numerical  illustrations. 

331.  1.  Change  the  proportion  a:b  =  c:  d  into  an  integral 
equation  by  Prin.  1. 

2.  Divide  the  members  of  this  equation,  be  =  ad,  by  ac,  and 
reduce. 

3.  What  change  has  taken  place  in  the  order  of  the  terms  of 
each  couplet  ? 

Principle  5.  —  If  four  numbers  are  in  proportion,  the  ratio 
of  the  second  to  the  first  is  equal  to  the  ratio  of  the  fourth  to  the 
third  ;  that  is,  the  numbers  are  in  proportion  by  Inversion. 

K  a:h  =  c:d,  then,  h  :  a  =  d  :  c. 

Demonstrate  Prin.  5,  and  give  numerical  illustrations. 

332.  1.  Express  the  proportion  a:b  =  c:d  as  a  fractional 
equation. 

2.  Add  1  to  each  member  and  reduce  the  mixed  numbers  to 
fractional  form.     Write  in  the  form  of  a  proportion. 

3.  How  may  the  terms  of  this  proportion  be  formed  from  the 
terms  of  the  given  proportion  ? 

4.  Since,  when  a:b  =  c:d,b:a  =  d:c,  if  the  changes  just 
indicated  are  made  in  the  second  proportion,  how  may  the  terms 
of  the  resulting  proportion  be  obtained  from  the  terms  of  the 
original  proportion? 


RATIO  AND  PROPORTION  331 

Principle  6.  —  If  four  numbers  are  in  proportion,  the  sum 
of  the  terms  of  the  first  ratio  is  to  either  term  of  the  first  ratio  as 
the  sum  of  the  terms  of  the  second  ratio  is  to  the  corresponding 
term  of  the  second  ratio  ;  that  is,  the  numbers  are  in  proportion  by 
Composition. 

If  a  lb  =  c  :d,  then,  a-{-b:b=:c  +  d:d  and  a  +  b  :  a  =  c  +  d :  c. 

Demonstrate  Prin.  6,  and  give  numerical  illustrations. 

333.  1.  Express  the  proportion  a:b  =  c:d  as  a  fractional 
equation. 

2.  Subtract  1  from  each  member,  and  reduce  the  mixed  num- 
bers to  fractional  form.     Write  in  the  form  of  a  proportion. 

3.  How  may  the  terms  of  this  proportion  be  formed  from  the 
terms  of  the  given  proportion  ? 

4.  Since,  Avhen  a  :  b  —  c :  d,  b  :  a  =  d  :  c,  if  the  changes  just 
indicated  are  made  in  the  second  proportion,  how  may  the  terms 
of  the  resulting  proportion  be  obtained  from  the  terms  of  the 
original  proportion  ? 

Principle  7.  —  If  four  numbers  are  in  j^^'oportion,  the  differ- 
ence between  the  terms  of  the  first  ratio  is  to  either  term  of  the 
first  ratio  as  the  difference  between  the  terms  of  the  second  ratio  is 
to  the  corresponding  term  of  the  second  ratio;  that  is,  the  numbers 
are  in  proportion  by  Division. 

If  a  :  6  =  c  :  (?,  then,  a  —  b:b  =  c  —  d:d  and  a  —  b  •.a  =  c  —  d-.c. 

Demonstrate  Prin.  7,  and  give  numerical  illustrations. 

334.  1.  Change  the  proportion  a:b  =  c:d  according  to  Prin. 
6,  and  also  according  to  Prin.  7,  using  the  same  consequents  in 
each  transformation.     Express  in  fractional  form. 

2.  Divide  the  first  equation  by  the  second. 

3.  How  may  the  terms  of  the  resulting  proportion  be  formed 
from  the  terms  of  the  given  proportion  ? 

Principle  8.  —  If  four  numbers  are  in  proportion,  the  sum  of 
the  terms  of  the  first  ratio  is  to  their  difference  as  the  sum  of  the 


332  RATIO   AND  PROPORTION 

terms  of  the  second  ratio  is  to  their  difference ;  that  is,  the  numbers 
are  in  proportion  by  Composition  and  Division. 

If  a  :  b=c  :  d,  a-\-b  :  a  —  b  —  c+d  :  c  —  d  and  a  +  b  :  b  —  a=c+d  :  d—c. 

Demonstrate  Prin.  8,  and  give  numerical  illustrations. 

335.  1.  Express  the  proportion  a:b  =  c:d  as  a  fractional 
equation. 

2.  Raise  each  member  to  the  nth  power. 

3.  Also  express  the  nth.  root  of  each  member. 

4.  How  may  these  proportions  be  formed  from  the  given 
proportion  ? 

Principle  9.  —  If  four  numbers  are  in  proportion,  their  like 

powers,  and  also  their  like  roots,  tvill  be  in  proportion. 

11       11 
If  a  :b  =  c  :d,  then,  a"  :  6"  =  c»  :  d"  and  a"  :  6"  =  c" :  d". 

Demonstrate  Prin.  9,  and  give  numerical  illustrations. 

336.  1.  Express  a  :  6  =  c :  d  as  a  fractional  equation. 

2.  What  may  be  done  to  a  fraction  without  changing  its  value  ? 

3.  Multiply  the  terms  of  the  first  fraction  by  m  and  the  terms 
of  the  second  fraction  by  n.     Write  as  a  proportion. 

4.  How  may  the  terms  of  this  proportion  be  formed  from  the 
terms  of  the  given  proportion  ? 

5.  Take  the  given  proportion  by  alternation  and  multiply  the 
terms  of  the  first  couplet  by  m  and  those  of  the  second  couplet 
by  w. 

6.  How  may  the  terms  of  this  proportion  be  formed  from  the 
terms  of  the  given  proportion  ? 

7.  How  may  the  given  proportion  be  formed  from  the  propor- 
tions ma  :  mb  =  nc:  nd  and  ma  :  nb  =  mc :  nd  ? 

Principle  10.  —  In  a  proportion,  if  both  terms  of  a  couplet,  or 
both  antecedents,  or  both  consequents  are  multiplied  or  divided  by 
the  same  number,  the  resulting  four  numbers  form  a  proportion. 

If  a  :  b  =  c  :  d,  then,  ma  :  mb  —  nc:  nd  and  ma  :  nb  =  mc  :  nd ;  also,  If 
ma  :  mb  =  nc  :  nd,  or  if  ma  :  nb  =  mc:  nd,  then,  a  -.b  =  c:  d. 

Demonstrate  Prin.  10,  and  give  numerical  illustrations. 


RATIO  AND  PROPORTION  333 

337.  1.  Express  the  proportions  a:b  =  c:d  and  x:y  =  z:w 
as  fractional  equations. 

2.  How  may  two  equations  be  combined  (Ax.  4  and  5)  ?  Com- 
bine these  two  equations  and  write  the  results  as  proportions. 

3.  How  may  these  proportions  be  formed  from  the  given 
proportions  ? 

Principle  11.  —  The  products,  and  also  the  quotients,  of  corre- 
sponding terms  of  two  proportions  form  a  proportion. 

K  a  -.b  =  c  :d  and  x  :  y  =  z  :  lo,  ax  :by  =  cz  :  dw,  also  _:__  =  _:_. 

X    y      z    w 

Demonstrate  Prin.  11,  and  give  numerical  illustrations. 

338.  If  a:h  =  c:d  and  c:d  =  e:f  how  does  the  ratio  a :  b 
compare  in  value  with  the  ratio  e :  f? 

Principle  12.  —  If  two  proportions  have  a  common  couplet,  the 
other  tivo  couplets  will  form  a  proportion. 

If  a  -.6  =  0  :d  and  c  :d  =  e  if,  then,  a  :  ft  =  e  : /. 

Demonstrate  Prin.  12,  and  give  numerical  illustrations  (Ax.  1). 

339.  A  proportion  that  consists  of  three  or  more  equal  ratios 
is  called  a  Multiple  Proportion. 

2:4  =  3:6  =  5:  10  and  a  .b  =  c  :d  =  e  -.f  are  multiple  proportions. 

340.  A  multiple  proportion  in  which  each  consequent  serves 
also  as  the  antecedent  of  the  following  ratio  is  called  a  Continued 
Proportion. 

2:4  =  4:8  =  8: 16  and  a  :b  =  b  :c  =  c:  d  are  continued  proportions. 

341.  1.  Form  a  multiple  proportion,  as 

2  :  4  =  3  :  6  =  5  :  10  =  10  :  20. 

2.  How  does  the  ratio  of  the  sum  of  the  antecedents  to  the  sum 
of  the  consequents  compare  with  the  first  ratio  ?  with  the  second 
ratio  ?   with  the  ratio  of  any  antecedent  to  its  consequent  ? 

3.  Investigate  other  multiple  proportions. 


334  RATIO  AND  PROPORTION 

Principle  13.  —  In  any  multiple  proportion  the  sum  of  the  ante- 
cedents is  to  the  sum  of  the  consequents  as  any  antecedent  is  to  its 
consequent. 

Principle  13  may  be  established  as  follows : 

Let  a  :b  =  c:d  =  e  :/=  g  :  h. 

It  is  to  be  proved  that 

a  +  c  +  e  +  g:b-{-d  +  f+h  =  a:b,  or  c:d,  etc. 
Denoting  each  of  the  equal  ratios  by  r, 

l  =  r,  |  =  r,   l=r,   l  =  r.  (1) 

b  d  f  h  ^  ^ 

Hence,  a  =  br,   c  =  dr,   e—fr,  g  =  hr.  (2) 

Adding  equations  (2), 

a  +  c  +  e  +  g=(b  +  d  +  f+h)r.  (3) 

Dividing  hy  (b  +  d  +  f+ h), 

a  +  c+e  +  g  ^^ 
b+d+f+h 

Replacing  r  by  any  of  the  equal  ratios, 

a  +  c  +  e  +  g  ^a_c   ^^^ 
b  +  d+f+h     b     d' 

That  is,     a  +  c  +  e  +  g  :b  +  d+f+  h  =  a:b,  OT  c-.d,  eta. 

Examples 
342.    1.  In  the  proportion  3  :  5  =  x :  55,  what  is  the  value  of  a;? 

First  Solution 
3  :  5  =  X  :  56. 

Prin.  2,  x=  ^^  =  33. 

5 

Second  Solution 
3  :  5  =  a; :  55. 
Prin.  10,  3  : 1  =  a; :  11, 

Prin.  1,  x=  33. 

Find  the  value  of  x  in  each  of  the  following  proportions : 

2.  2:3  =  6  :a;.  4.    l:a;  =  a;:9. 

3.  5:«  =  4:3.  5.    8:5  =  x:10. 


RATIO   AND  PROPORTION  335 

6.  «  +  1  .  a;  =  8  :  6.  8.    a;  +  2  :  a;  =  10  :  6. 

7.  a;:  a;- 1  =  15:  12.  9.    a;  + 2  :  a;- 2  =  3  : 1. 
I     10.    If  a;  +  5  :  a;  —  5  =  5  :  3,  find  the  value  of  x. 

11.  What  two   numbers   are   mean   proportionals   between   1 
and  25? 

12.  Show  that  a  mean  proportional  between  any  two  numbers 
has  the  sign  ± . 

When  a  :  b  =  c  :  d,  prove  that  the  following  proportions   are 
true  by  deriving  them  from  a:b  =  c:  d: 

13.  d:b  =  c.a.  16.    a^ :  6V  =  1  :  d^ 

,.„,11  __  &  d 

14.  c:a  =  -:—  17.    ma  :  -  =  mc  :  ~' 

b   a  2  2 

15.  b^ :  d^  ^  a^ :  c^.  18.    ac  :  bd  =  c^ :  d". 

19.  Vod  :  V&  =  Vc  :  1. 

20.  a  +  b  :  c  -\-  d  =  a  —  b  :  c  —  d.  p 

21.  a"  +  a%  +  ab"^  -(-  b^ :  a^  =  <? ->^  <?d  -\-cd^  ■\-d^:<?.    ^  ' 

22.  2a  +  3c:2a-3c  =  86  +  12d:86-12rf. 

23.  Solve  the  equation  ^'  +  ^  =  — ■ 

^  oa;         20 

Solution 

gg  +  x2  _  41 
ax         20' 

Dividing  by  2,  «i±^  =  il. 

2  ax        40 

Regarding  this  equation  as  a  proportion, 

by  composition  and  division,  -    +  ^  «^  +  ^g  _  81^ 
a^  —  2  ax  4-  x'^      1 

-  Extracting  the  square  root,      '^         =  — 

a  —  X     1 

composition  and  division,       —  =  — . 
2x      8 

.-.  x  =  |a. 


336  RATIO  AND  PROPORTION 


24.    Solve  the  equation 


Vaj  +  T+VcB      Va;  +  7  - ^^ 


4+Va 


Solution 


■■\x 


y/x+1  +^/x _y/x+l  -v'x 
4  +  Va;  4  —  y/x 


By  alternation,  Prin.  4,     ^^^^^+^  =  ^  +  ^^. 
v^a;  +  7  -  Vx     4  -  Vx 

By  composition  and  division,  Prin.  8, 


2Vx+7  _    8 
2Vx         2Vx 
Since  the  consequents  are  equal,  the  antecedents  are  equal. 
Therefore,  2\/x  +  7  =  8. 

Whence,  reducing,  x  =  9. 


25.    Given  V^' +  11  +  2^  V2^+14  +  2|_  ^^  ^^^  ^ 
Vaj  +  11  -  2     V2  aj  +  14  -  2| 


Solution 
Vx  +  11  +  2  _  ^2  X  +  14  + 


v'x+11-2      V2x-fl4 
By  composition  and  division,  Prin.  8, 


2Vx  +  ll_2V2x  +  14 
4  Y 

Dividing  both  terms  of  each  ratio  by  2,  Prin.  10, 


Vx  +  11_  \/2x+  14 

Dividing  the  consequents  by  f ,  Prin.  10, 

Vx  +  ll_  V2X+14 
3  4 


By  alternation,  Prin.  4,  ^x,  +  U    _  3_ 

V2  X  +  14     4 


Squaring,  and  applying  Prin.  7, 


x  +  ll_  9 


x  +  3      7 
Solving,  X  =  25. 


RATIO   AND  PROPORTION  337 


Solve  by  the  principles  of  proportion ; 

ViB  —  Vm      ^ 

V^-fV2a_2 


26.    V^±V^  =  ^.  29. 


27.    V|W2a^2.  3^_ 


Va; 

+  6  +  Va;  - 

-&_ 

-  d. 

Vx 

+  6- 

4-Va 

Vx- 

1 

a 

IX  — 

Va 

—  X 

Vet  —  Va 
Voa;  —  & 

—  a; 
3V( 

26 

28.    ^  +  V^^13.  3^ 

a;  —  Va;  —  1       ^  ^aa;  +  h      3  Vaa;  +  56 

„-     ^  ■         Va;  +  2  +  Vi  +  1      3   ,     r;    , 

32.  Given         ^     ^        ^     =  -,  to  find  x. 

Va;  +  2  +  Va;  —  1      1 

_„     ^ .        Va  4-  Va  +  X     V6  4-  Va;  —  6  .     n    •, 

33.  Given  '         ^     =        ^  to  find  a;. 

Va  —  Va  +  a;      V6  —  Va;  —  6 


/^-        a  —  V2  ax  —  x^     a  —  b   .     n    ■, 
34.    Given —  = ,  to  find  x. 

a+V2aa;-a;2     a  +  b 


„,,     />.•         Va;  +  1  +  Va;  —  2      Va;  —  3  +  -v'a;  —  4  .    ^    i 

35.  Given  ^       '  =  '  ,  to  find  x. 

Va;  +  1  —  Va;  —  2      Va;  —  3  —  Va;  —  4 

36.  Divide  $  35  between  two  men  so  that  their  shares  shall  be 
in  the  ratio  of  3  to  4. 

37.  Two  numbers  are  in  the  ratio  of  3  to  2,  and  if  each  is 
increased  by  4,  the  sums  will  be  in  the  ratio  of  4  to  3.  What  are 
the  numbers  ? 

38.  Divide  16  into  two  parts  such  that  their  product  is  to  the 
sum  of  their  squares  as  3  is  to  10. 

39.  Divide  25  into  two  parts  such  that  the  greater  increased 
by  1  is  to  the  less  decreased  by  1  as  4  is  to  1. 

40.  The  sum  of  two  numbers  is  4,  and  the  square  of  their  sum 
is  to  the  sum  of  their  squares  as  8  is  to  5.     What  are  the  numbers  ? 

41.  A  dealer  had  two  casks  of  wine.  From  the  larger  he  drew 
34  gallons,  and  from  the  smaller  8  gallons,  after  which  their  con- 
tents were  as  5  to  4.  When  half  the  original  contents  of  each 
cask  had  been  drawn,  he  put  8  gallons  into  the  larger  and  6  into 
the  smaller.  If  the  ratio  of  their  contents  was  then  5  to  3,  what 
was  the  capacity  of  each  ? 

ALG.  —22 


VARIATION 


343.  One  quantity  or  number  is  said  to  vary  directly  as  another, 
or  simply  to  vary  as  another,  when  they  depend  on  each  other  in 
such  a  manner  that  if  one  is  changed  the  other  is  changed  in  the 
same  ratio. 

Thus,  if  a  man  earns  a  certain  sum  per  day,  the  amount  of  wages  he  earns 
varies  as  the  number  of  days  he  works. 

344.  The  Sign  of  Variation  is  cc.     It  is  read  '  varies  as.' 

345.  The  expression  xc^y  means  that  if  x  is  doubled,  y  is 
doubled,  or  if  x  is  divided  by  a  number,  y  is  divided  by  the  same 
number,  etc. ;  that  is,  that  the  ratio  of  cc  to  3/  is  always  the  same, 
or  constant. 

If  the  constant  ratio  is  represented  by  k,  then  when  ccocy, 
-  =  A;,  or  x  =  ky.     Hence, 

y 

If  x  varies  as  y,  x  is  equal  to  y  multiplied  by  a  constant. 

346.  One  quantity  or  number  varies  inversely  as  another  when 
it  varies  as  the  reciprocal  of  the  other. 

Thus,  the  time  required  to  do  a  certain  piece  of  work  varies  inversely  as 
the  number  of  men  employed.  For,  if  it  takes  10  men  4  days  to  do  a  piece  of 
work,  it  will  take  5  men  8  days,  or  4  men  10  days,  to  do  it. 

1    .  1  X 

In  a;  oc  -,  if  the  constant  ratio  of  cc  to  -  is  Zc,  t  =  ^»  or  aw  =  Tc. 

Hence,  y 

If  X  varies  inversely  as  y,  their  product  is  constant. 

347.  One  quantity  or  number  varies  jointly  as  two  others  when 
it  varies  as  their  product. 

888 


VARIATION  339 

Thus,  the  amount  of  money  a  man  earns  varies  jointly  as  the  number  of 
days  he  works  and  the  sum  he  receives  per  day.  For,  if  he  should  work 
three  times  as  many  days,  and  receive  twice  as  many  dollars  per  day,  he 
would  receive  six  times  as  much  money. 

In  a:  oc  yz,  if  the  constant  ratio  of  x  to  yz  is  A:, 

—  ='k,  or  X  —  Jcyz.     Hence, 
yz 

If  X  varies  jointly  as  y  and  z,  x  is  equal  to  their  prodrict  multiplied 
by  a  constant. 

348.  One  quantity  or  number  varies  directly  as  a  second  and 
inversely  as  a  third  when  it  varies  jointly  as  the  second  and  the 
reciprocal  of  the  third. 

Thus,  the  time  required  to  dig  a  ditch  varies  directly  as  the  length  of  the 
ditch  and  inversely  as  the  number  of  men  employed.  For,  if  the  ditch  were 
10  times  as  long  and  5  times  as  many  men  were  employed,  it  would  take 
twice  as  long  to  dig  it. 

1  V 

In  xccy  '  -,  or  xcc~,  if  Z;  is  the  constant  ratio, 
z  z  ' 

x-^'-  =  ]c,  or  x  =  k--     Hence, 

z  z  ' 

If  x  varies  directly  as  y  and  inversely  as  z,  x  is  equal  to  ^  multi- 
plied by  a  constant. 

349.  If  X  varies  as  y  when  z  is  constant,  and  x  varies  as  z  when 
y  is  constant,  then  x  varies  as  yz  lohen  both  y  and  z  are  variable. 

Proof.  —  Since  the  variation  of  x  depends  on  the  variations  of  y  and  z, 
suppose  the  latter  variations  to  take  place  in  succession,  each  in  turn  pro- 
ducing a  corresponding  variation  in  x. 

While  z  remains  constant,  let  y  change  to  yi*  thus  causing  x  to  change 
to  x'. 

Then,  ^=y-.  (1) 

a;'     2/1 

Now  while  y  keeps  the  value  yi,  let  z  change  to  «i,  thus  causing  x'  to 
change  to  x\. 

Then,  *!  =  £.  (2) 

Xi       Zi 

*  In  algebraic  notation  Xi,  X2,  Xs,  etc.,  are  read  'a;  sub  one,'  'a;  sub  two,' 
'X  sub  three,'  etc. 


340 

Multiplying  (1)  by  (2), 


VARIATION 

X 

_  yz 

Xi' 

yizi 

:.  X  : 

Xl 

yizi 

yz 

kyz, 


^1 

where  k  is  the  constant 

yizi 
Hence,  x  cc  yz. 

Thus,  the  area  of  a  triangle  varies  as  the  base  when  the  altitude  is  con- 
stant, varies  as  the  altitude  when  the  base  is  constant,  and  varies  as  the 
product  of  the  base  and  altitude  when  both  vary. 

Similarly,  if  x  varies  as  each  of  three  or  more  numbers,  y,  z, 
V,  •••  when  the  others  are  constant,  when  all  vary  x  varies  as  their 
product. 

That  is,  xccyzv  ••-. 

Thus,  the  volume  of  a  parallelopiped  varies  as  the  length,  if  the  width  and 
thickness  are  constant ;  as  the  width,  if  the  length  and  thickness  are  con- 
stant ;  as  the  thickness,  if  the  length  and  width  are  constant ;  as  the  product 
of  any  two  dimensions,  if  the  other  is  constant ;  or  as  the  product  of  the 
three  dimeusions,  if  all  vary. 

ExAMPIiES 

350.  1.  li  X  varies  inversely  as  y,  and  x  =  6  when  y  =  S,  what 
is  the  value  of  x  when  y  —  12? 

Solution 
Since  x  x  -,  let  k  be  the  constant  ratio  of  x  to  — 

y  y 

Then,  §  346,  xy  =  k.  (1) 

Hence,  when  x  =  6  and  y  =  8, 

A;  =  6  X  8,  or  48.  (2) 

Since  k  is  constant.  A;  =  48  when  y  =  12. 

Hence,  Eq.  (1)  becomes  12  a;  =  48. 

Therefore,  when  y  —  \2,  x  =  i. 

2.  The  volume  of  a  cone  varies  jointly  as  its  altitude  and  the 
square  of  the  aiameter  of  its  base.  When  the  altitude  is  15  and 
the  diameter  of  the  base  is  10,  the  volume  is  392.7.  What  is  the 
volume,  when  the  altitude  is  5  and  the  diameter  of  the  base 
is  20? 


VARIATION  341 

•  '  Solution 

Let  V,  H,  and  D  denote  the  volume,  altitude,  and  diameter  of  the  base, 
respectively,  of  any  cone,  and  v  the  volume  of  a  cone  whose  altitude  is 

5  and  the  diameter  of  whose  base  is  20. 

Since  VxHD^,  or  V=kHD^, 

and  V=  392.7  when  H=lo  and  D  =  10, 

392.7  =  A;  X  15  X  100.  (1) 

Also,  since  V  becomes  v  when  H=b  and  D  =  20, 

v  =  kx  5x  400.  (2) 

Dividing  (2)  by  (1),      -^^  =  ^  ^  ^^  =  1  (3) 

.-.  -y  =  f  of  392.7  =  523.6. 

3.  It  xccy  and  yccz,  prove  that  xacz. 

Proof 

Since  xccy  and  y  oc  2,  let  m  represent  the  constant  ratio  of  x  to  y,  and 
n  the  constant  ratio  of  y  to  2. 

Then,  §345,  x  =  my,  (1) 

and  y  =  m.  (2) 

Substituting  n^  for  y  in  (1),       x  =  mm.  (3) 

Hence,  since  mn  is  constant,  xcc  z. 

4.  The  circumference  of  a  circle  varies  as  its  diameter.  If 
the  circumference  of  a  circle  whose  diameter  is  1  foot  is  3.1416 
feet,  what  is  the  circumference  of  a  circle  whose  diameter  is 
100  feet? 

5.  The  area  of  a  circle  varies  as  the  square  of  its  diameter. 
If  the  area  of  a  circle  whose  diameter  is  10  feet  is  78.54  square 
feet,  what  is  the  area  of  a  circle  whose  diameter  is  20  feet  ? 

6.  The  distance  a  body  falls  from  rest  varies  as  the  square 
of  the  time  of  falling.  If  a  stone  falls  64.32  feet  in  2  seconds, 
how  far  will  it  fall  in  5  seconds  ? 

7.  The  area  of  a  triangle  varies  jointly  as  its  base  and  alti- 
tude.    The  area  of  a  triangle  whose  base  is  12  inches  and  altitude 

6  inches  is  36  square  inches.  What  is  the  area  of  a  triangle 
whose  base  is  8  inches  and  altitude  10  inches?  What  is  the 
constant  ratio  ? 


342  VA  RIA  TION 

8.  A  wrought  iron  bar  1  square  inch  in  cross  section  -and 
1  yard  long  weighs  10  pounds.  If  the  weight  of  a  uniform  bar 
of  given  material  varies  jointly  as  its  length  and  the  area  of  its 
cross  section,  what  is  the  weight  of  a  wrought  iron  bar  36  feet 
long,  4  inches  wide,  and  4  inches  thick  ? 

9.  The  weight  of  a  beam  varies  jointly  as  the  length,  the  area 
of  the  cross  section,  and  the  material  of  which  it  is  composed. 
If  wood  is  Y^<j  as  heavy  as  wrought  iron,  what  is  the  weight  of  a 
wooden  beam  24  feet  long,  12  inches  wide,  and  12  inches  thick  ? 

10.  What  is  the  weight  of  a  brick  2  in.  x  4  in.  x  8  in.,  if  the 
material  is  \  as  heavy  as  wrought  iron  ? 

11.  If  10  men  can  do  a  piece  of  work  in  20  days,  how  long  will 
it  take  25  men  to  do  it  ? 

12.  If  a  men  can  do  a  piece  of  work  in  b  days,  how  many  men 
will  be  required  to  do  it  in  c  days  ? 

13.  The  distances,  from  the  fulcrum  of  a  lever,  of  two  weights 
that  balance  each  other  vary  inversely  as  the  weights.  If  two 
boys  weighing  80  and  90  pounds,  respectively,  are  balanced  on  the 
ends  of  a  board  8|^  feet  long,  how  much  of  the  board  has  each  ? 

14.  A  water  carrier  carries  two  buckets  of  water  suspended 
from  the  ends  of  a  4-foot  stick  that  rests  on  his  shoulder.  If 
one  bucket  weighs  60  pounds  and  the  other  100  pounds,  and  they 
balance  each  other,  what  point  of  the  stick  rests  on  his  shoulder  ? 

15.  The  weight  of  a  body  near  the  earth  varies  inversely  as 
the  square  of  its  distance  from  the  center  of  the  earth.  If  the 
radius  of  the  earth  is  4000  miles,  what  would  be  the  weight  of  a 
4-lb.  brick  at  the  distance  of  4000  miles  from  the  earth's  surface  ? 

16.  A  boy  wishes  to  ascertain  the  height  of  a  tower.  He 
knows  that  it  is  31  feet  6  inches  from  his  window  to  the  pave- 
ment below,  and  that  the  distance  through  which  a  body  falls 
varies  as  the  square  of  the  time  of  falling.  He  drops  a  marble 
from  his  window  and  finds  that  it  strikes  the  pavement  in  1.4 
seconds.  Then  he  throws  a  stone  to  the  top  of  the  tower  and 
observes  that  it  takes  just  3  seconds  for  it  to  descend.  What 
is  the  height  of  the  tower  ? 


VARIATION  343 

17.  A  horse  tied  with  a  rope  45  feet  long  to  a  stake  in  the 
center  of  a  pasture  eats  all  the  grass  within  reach  in  1^  days.  If 
his  rope  were  15  feet  longer,  how  many  days  would  it  take  hira  to 
eat  all  the  grass  within  reach  ? 

18.  The  illumination  from  a  source  of  light  varies  inversely 
as  the  square  of  the  distance.  How  far  must  a  screen  that  is  10 
feet  from  a  lantern  be  moved  so  as  to  receive  one  fourth  as  much 
light  ? 

19.  The  number  of  times  a  pendulum  oscillates  in  a  given 
time"  varies  inversely  as  the  square  root  of  its  length.  If  a  pen- 
dulum 39.1  inches  long  oscillates  once  a  second,  what  is  the 
length  of  a  pendulum  that  oscillates  twice  a  second? 

20.  How  long  must  a  pendulum  be  to  oscillate  once  in  three 
seconds  ? 

II 

21.  If  XCC-,  and  if  x  =  2  when  y  =  12,  and  2  =  2,  what  is  the 

value  of  X  when  ?/  =  84  and  z  =  l  ? 

22.  If  XCC-,  and  if  aj  =  60  when  y  =  24,  and  2  =  2,  what  is  the 
value  of  y  when  cc  =  20  and  2  =  7? 

23.  If  X  varies  jointly  as  y  and  z  and  inversely  as  the  square 
of  w,  and  if  a;  =  30  when  ?/  =  3,  2  =  5,  and  to  =  4,  what  is  the 
value  of  X  expressed  in  terms  of  y,  z,  and  w? 

24.  If  ic«-  and  ycc-,  prove  that  xccz. 

25.  li  xccy  and  zccy,  prove  that  (x  ±  z)xy. 

26.  Three  spheres  of  lead  whose  radii  are  6,  8,  and  10  in.,  re- 
spectively, are  united  into  one.  Find  the  radius  of  the  resulting 
sphere,  if  the  volume  of  a  sphere  varies  as  the  cube  of  its  radius. 

27.  The  volume  of  a  cone  varies  jointly  as  its  altitude  and  the 
square  of  the  diameter  of  its  base.  The  altitudes  of  three  cones, 
8,  P,  and  E,  are  30  ft.,  10  ft.,  and  5  ft.,  respectively.  The 
diameter  of  the  base  of  P  is  5  ft.  and  that  of  E  is  10  ft.  If  the 
volume  of  S  is  equivalent  to  that  of  P  and  E  combined,  what  is 
the  diameter  of  the  base  of  ^S  ? 


PROGRESSIONS 


351.  1.  How  does  each  of  the  numbers  2,  4,  6,  8,  10,  12,  ••• 

compare  with  the  number  that  follows  it  ?     How  may  any  term 
after  the  first  be  obtained  from  the  preceding  term  ? 

2.  How  may  any  term  of  2,  2^,  3,  Z\,  •••  after  the  first  be  ob- 
tained from  the  preceding  term  ? 

3.  Write  a  series  of  six  terms  beginning  with  a  and  increasing 
by  a  constant  number  d. 

4.  How  may  any  term,  after  the  first,  of  the  series  3,  6,  12, 
24,  48,  •••be  obtained  from  the  preceding  term  ? 

5.  How  may  any  term,  after  the  first,  of  the  series  1,  ^,  \,  \,  ••• 
be  obtained  from  the  preceding  term  ? 

6.  Write  a  series  of  six  terms  beginning  with  a  and  increasing 
by  a  constant  multiplier  r. 

352.  A  succession  of  numbers,  each  of  which  after  the  first  is 
derived  from  the  preceding  number  or  numbers  according  to  some 
fixed  law,  is  called  a  Series. 

353.  The  successive  numbers  are  called  the  Terms  of  the  series. 
The  first  and  last  terms  of  a  series  are  the  Extremes,  the  inter- 
vening terms  the  Means. 

In  the  series  a,  a  +  d,  a  +  2  d,  c  +  3 (^,  a  +  4 d,  the  terms  a  and  a  +  id 
are  the  extremes  and  the  other  terms  are  the  means. 

354.  A  series  consisting  of  a  limited  number  of  terms  is 
called  a  Finite  Series. 

355.  A  series  consisting  of  an  unlimited  number  of  terms  is 
called  an  Infinite  Series. 

344 


PROGRESSIONS  345 

ARITHMETICAL  PROGRESSION 

356.  A  series  each  term  of  which  after  the  first  is  den  ^ed  from 
the  preceding  by  the  addition  of  a  constant  number  is  called  an 
Arithmetical  Series,  or  an  Arithmetical  Progression. 

357.  The  number  added  to  any  term  to  produce  the  next  is 
called  the  Common  Difference. 

2,  4,  6,  8,  •••  and  15,  12,  9,  6,  •••  are  arithmetical  progressions.  In  the 
first,  the  common  difference  is  2  and  the  series  is  ascending ;  in  the  second, 
the  common  difference  is  —  3  and  the  series  is  descending. 

A.  P.  is  an  abbreviation  of  the  word.s  Arithmetical  Progression. 

358.  To  find  the  nth,  or  last  term. 

1.  In  the  arithmetical  progression  a;,  cc  +  2,  cc  +  4,  a;  +  6,  what 
is  the  common  difference  ?  How  many  times  does  it  enter  into 
the  second  term  ?  into  the  third  term  ?  into  the  fourth  term  ? 

2.  From  the  first  term  of  the  series  a,  a  +  d,  a  +  2  d,  a  +  ^  d,  '•' 
how  is  the  second  term  formed  ?  the  third  term  ?  the  fourth 
term  ?  the  fifth  term  ?  the  nth  term,  or  any  term  ? 

3.  What  is  the  nth.  term  of  the  series  a,  a  —  d,  a  —  2  d,  ••■? 

369.  When  a  represents  the  first  term  of  an  A.P.,  d  the  com- 
mon  difference,  I  the  nth,  or  last  term,  and  n  the  number  of  terms, 

l  =  a  +  (n~l)d.  (I) 

Examples 

1.   What  is  the  10th  term  of  the  series  3,  6,  9,  •••  ? 

PROCESS  Explanation. —Since  the  series  3,  6,  9,  •••  is  an 

Z  =  a  -f  (w  —  l)d  A. P.  the  common  difference  of  whose  terms  is  3,  sub- 

^  _  3  ^  (10 1)3  stituting  3  for  «,  3  for  d,  and  10  for  n  in  the  formula 

7  ^  oA  for  the  last  term,  the  last  term  is  found  to  be  30. 

2.  Find  the  20th  term  of  the  series  7,  11,  15,  •••. 

3.  Find  the  16th  term  of  the  series  2,  7,  12,  •••. 

4.  Find  the  24th  term  of  the  series  1,  16,  31,  ••♦. 


346  PROGRESSIONS 

5.  Find  the  18th  term  of  the  series  1,  8,  15,  •••. 

6.  Find  the  13th  term  of  the  series  —  3,  1,  5,  •••. 

7.  Find  the  49th  term  of  the  series  1,  1|,  If,  •••. 

8.  Find  the  15th  term  of  the  series  45,  43,  41,  •••. 
Suggestion.  —  The  common  difference  is  —  2. 

9.  Find  the  10th  term  of  the  series  5,  1,  —  3,  •••. 

10.  Find  the  16th  term  of  the  series  a,  3  a,  5  a,  •••. 

11.  Find  the  12th  term  of  the  series  a  —  b,  a  +  6,  a  +  3b,  •••. 

12.  Find  the  7th  term  of  the  series  x  ~  3y,  x  —  2 y,  •••. 

13.  A  body  falls  16^1^  feet  the  first  second,  3  times  as  far  the 
second  second,  5  times  as  far  the  third  second,  etc.  How  far  will 
it  fall  during  the  10th  second  ? 

360.  To  find  the  sum  of  n  terms  of  a  series. 

1.  Express  5  terms  of  the  series  a,  a  -\-  d,  a  -{-2d,  •••. 

2.  How  may  the  term  before  the  last  term  be  obtained  from 
the  last  term  ?  If  I  represents  the  last  term  and  d  the  common 
difference,  what  will  be  the  term  next  to  the  last  ?  the  second 
term  from  the  last  ?  the  third  term  from  the  last  ? 

3.  How,  then,  may  the  series  a,  a  +  d,  •••  be  written  in  reverse 
order,  if  the  last  term  is  I? 

361.  Let  a  represent  the  first  term  of  an  A.P.,  d  the  common 
diiference,  /  the  last  term,  n  the  number  of  terms,  and  s  the  sum 
of  the  terftis. 

Writing  the  sum  of  n  terms  in  the  usual  order  and  then  in  the 
reverse  order,  and  adding  the  two  equal  series, 

s  =  tt  +  (a  +  d)  +  (a  +  2  d)  +  (a  +  3  d)  H +1 

s  =  I  -\-  (I  -  d)  +  (I  -  2  d)  +  (I  -  3  d)  +  -■  +  a. 

2  s  =  (a  +  0  +  (a  +  0  +  (a  +  0  +  (a  +  0  +  •••  +  («  +  0- 
.\2s  =  n(a-\-l). 

s='^{a  +  l),ovn(^'^y  (II) 


PROGRESSIONS 


347 


Examples 


1.    What  is  the  sum  of  20  terms  of  the  series  2,  5,  8,  •••  ? 

PROCESS 

2  =  a  +  (w  -  1)  d  =  ^  +  (20  -  1)  X  3  =  59 


a-\-l 


2   y        V    2    ; 


-±^^= 


610 


Explanation,  —  Since  the  last  term  is  not  given,  it  is  found  by  the  pre- 
vious case  and  substituted  for  I  in  the  formula  for  the  sum. 


2.  What 

3.  What 

4.  What 

5.  What 

6.  WTiat 

7.  What 

8.  What 

9.  What 
10.  What 


s  the  sum  of  16  terms  of  the  series  1,  5,  9,  •••  ? 
s  the  sum  of  10  terms  of  the  series  —  2,  0,  L',  •••  ? 
s  the  sum  of  6  terms  of  the  series  1,  3^,  6,  •••  ? 
s  the  sum  of  8  terms  of  the  series  a,  3  a,  5  a,  •••  ? 
s  the  sum  of  n  terms  of  the  series  1,  7,  13,  •••  ? 
s  the  sum  of  a  terms  of  the  series  x,  x  -\-  2  a,  •••  ? 
s  the  sum  of  7  terms  of  the  series  4, 11,  18,  •••  ? 
s  the  sum  of  10  terms  of  the  series  1,  —  1,  —  3,  ••• 
s  the  sum  of  10  terms  of  the  series  1,  |,  0,  •••  ? 


11.  How  many  times  does  a  common  clock  strike  in  12  hours  ? 

12.  A  body  falls  16^  feet  the  first  second,  3  times  as  far  the 
second  second,  5  times  as  far  the  third  second,  etc.  Hbw  far  will 
it  fall  in  10  seconds  ? 

13.  Thirty  flower  pots  are  arranged  in  a  straight  line  4  feet 
apart.  How  far  must  a  lady  walk  who,  after  watering  each 
plant,  returns  to  a  well  4  feet  from  the  first  plant  and  in  line 
with  the  plants,  assiuning  that  she  starts  at  the  well  ? 

14.  A  boy  took  a  30-day  job  on  the  following  terms :  he  was 
to  receive  5  cents  the  first  day,  10  cents  the  second  day,  15  cents 
the  third  day,  etc.  How  much  was  he  paid  for  the  thirtieth  day, 
and  what  was  the  whole  amount  of  his  earnings  ? 


348  PROGRESSIONS 

362.    The  two  fundamental  formulae, 

(I)  l  =  a  +  (7i-l)d  and  (II)  s=-(a  +  0» 

Li 

contain  five  elements,  a,  d,  I,  n,  and  s.  Since  these  formulae  are 
independent  simultaneous  equations,  if  they  contain  but  two 
unknown  elements  they  may  be  solved.  Hence,  if  any  three  of 
the  five  elements  are  knoivn,  the  other  two  may  be  found. 


Examples 

1.  The  last  term  of  an  A.  P.  is  58,  the  common  difference  is  3, 
and  the  sum  of  the  series  is  260.  Find  the  number  of  terms  and 
the  first  term. 

Solution 

Substituting  58  for  Z,  3  for  d,  and  260  for  s  in  both  (I)  and  (II), 

(I)  becomes  58  =  a  +  (n  -  1)3.  (1) 

(II)  becomes  260  =  -  (a  +  58) .  (2) 

(1)  X  n,  58  n  =  na  +  3  w^  -  3  n.  (3) 

(2)  X  2,  520  =  na  +  58  n.  (4) 

(3)  -  (4) ,  58  n  -  520  =  3  n^  -  61  n.  (5) 

3n2  -  119  n  + 520  =  0. 

(n-5)(3n- 104)  =  0. 

.•.  «  =  5,  the  number  of  terms. 

Substituting  in  (1),  a  =  46,  the  first  term. 

Since  the  number  of  terms  must  be  expressed  by  a  positive  integer,  frac- 
tional or  negative  values  of  n  are  rejected. 

2.  How  many  terms  are  there  in  the  A.  P.  2,  5,  8,  •••,  if  the 
sum  is  610  ? 

Solution 

Since  a,  d,  and  s  are  given,  and  n,  but  not  I,  is  required,  n  may  be  found 
by  eliminating  I  from  (I)  and  (II)  and  solving  the  resulting  equation. 

From  (I)  and  (II),      l  =  a+(n-l)d  =  iLniHi. 


Substituting  2  for  c,  3  for  d,  and  610  for  s,  and  solving, 

n  =  20. 


PROGRESSIONS  349 

t3.    How  many  terms  are  there  in  the  series  2,  6,  10,  •••  66  ? 
4.    What  is  the  sum  of  the  series  1,  6,  11,  •••  61  ? 

5.  How  many  terms  are  there  in  the  series  —  1,  2,  5,  •••,  if 
the  sum  is  221  ? 

6.  Determine  the  series  2,  9,  16,  •••  86. 

7.  Determine  the  series  —  10,  —  8^,  —  7,  •••to  10  terms. 

8.  The  sum  of  the  series  •••22,  27,  32,  •••  is  714.     If  there 
ire  17  terms,  what  are  the  first  and  last  terms  ? 

9.  If  s  =  113§,  a  =  ^,  and  d  =  2,  find  n. 

10.  What  is  the  sum  of  the  series  -  16,  —  11,  —  6,  •••  34  ? 

11.  What  is  the  sum  of  the  series  •••  —1,  3,  7,  •••  23,  if  the 
number  of  terms  is  16  ? 

12.  What  are  the  extremes  of  the  series  •••  8,  10,  12,  •••,  if  s  = 
300,  and  n  =  20  ? 

13.  How  many  terms  are  there  in  the  series  1,  5,  9,  •••  V^ 

14.  What  is  the  sum  of  an  A.  P.  whose  extremes  are  x  and  y, 
if  the  number  of  terms  is  6  ? 

363.    To  insert  arithmetical  means. 

Examples 

1.  Insert  5  arithmetical  means  between  1  and  31. 

Solution.  —  Since  there  are  5  means,  there  must  be  7  terms.     Hence,  in 
I  =  a  -\- {n  —  \)  d,  Z  =  31,  a  —  \,  n  =  7,  and  d  is  unknown. 

Solving,  d  =  5. 

Or,  since  there  are  6  means,  there  must  be  6  terms  after  the  first. 

Hence,  d  =  5ijzJ  =  5. 

6 

.-.  1,  6,  11,  16,  21,  26,  31,  is  the  series. 

2.  Insert  9  arithmetical  means  between  1  and  6. 

3.  Insert  10  arithmetical  means  between  24  and  2. 

4.  Insert  7  arithmetical  means  between  10  and  —  14. 

5.  Insert  6  arithmetical  means  between  —  1  and  2. 


350  PROGRESSIONS 

6.  Insert  14  arithmetical  means  between  15  and  20. 

7.  Insert  3  arithmetical  means  between  a  —  h  and  a  +  6. 

8.  Deduce  the  formula  for  the  common  difference  when  m 
arithmetical  means  are  to  be  inserted  between  a  and  I.  Find  the 
first  mean. 

9.  What  is  the  arithmetical  mean  between  2  and  6  ?  between 
10  and  20  ?   between  —  3  and  5  ?   between  a  and  6  ? 

364.  Principle.  —  The  arithmetical  mean  between  two  numbers 
is  equal  to  half  their  sum. 

The  above  principle  may  be  established  as  follows : 

Let  a  and  b  represent  any  two  numbers,  and  A  their  arithmetical  mean. 

It  is  to  be  proved  that  A  =  ^  "*"  - 

Since  the  two  numbers  and  their  arithmetical  mean  form  the  arithmetical 
progression  a,  A,  b, 

§  356,  A  —  a  =  b  —  A, 

2A  =  a  +  b. 

.:  A  =  «i-^. 
2 

Examples 
Find  the  arithmetical  mean  between 

1-    fandi  ^    ^+Jand^^^. 

X  —  y  X  -{-  y 

2.  a  +  6  and  a  —  b. 

5.    1  -  a;  and  N-*-  ~  ^)  • 

3.  (a  +  6)2  and  (a  -  6)1  1  +  x 

Problems 

365.  Problems  in  Arithmetical  Progression  involving  two 
unknown  elements  commonly  suggest  series  of  the  form 

X,   x  +  y,   x  +  2y,   x  +  3y,   etc. 

Frequently,  however,  the  solution  of  problems  is  more  readily 
accomplished  by  representing  the  series  as  follows : 

1.  When  there  are  three  terms,  the  series  may  be  written 

x-y,   X,   x  +  y. 


PR0GRE»8I0NS  351 

2.  When  there  are  five  terms,  the  series  may  be  written, 

a;  -  2  2/,  x  -  y,  x,  x  +  y,  x  +  2  y. 

3.  "When  there  are  foxir  terms,  the  series  may  be  written, 

X  -  3  y,  X  -  2/,  X  +  J/,  X  +  3  y. 

The  sum  of  the  terms  of  a  series  represented  as  above  evidently  contains 
but  one  unknown  number. 

1.  The  sum  of  three  numbers  in  arithmetical  progression  is 
30,  and  the  sum  of  their  squares  is  462.     What  are  the  numbers  ? 

Solution 
Let  the  series  be  as  —  y,  x,  x  +  y. 

Then,  (x  -  y)  +  x  +  (x  +  y)  =  30,  (1) 

and  (X  -  i/)2  +  x2  +  (X  +  y)^  =  462.  (2) 

From  (1),  3x  =  30.  (3) 

X  =  10.  (4) 

From  (2),  3  x^  +  2  y2  =  462.  (5) 

Substituting  10  for  X,  2y2  =  162.    •  (6) 

Solving,  y  =  ±  9. 

Forming  the  series  from  x  =  10  and  y  =  ±  9,  the  terms  are 
1,  10,  19,  or  19,  10,  1. 

2.  The  sum  of  three  numbers  in  arithmetical  progression  is 
18,  and  their  product  is  120.     What  are  the  numbers  ? 

3.  The  sum  of  three  numbers  in  arithmetical  progression  is 
21,  and  the  sum  of  their  squares  is  155.     What  are  the  numbers  ? 

4.  There  are  three  numbers  in  arithmetical  progression  the 
sum  of  whose  squares  is  93.  If  the  third  is  4  times  as  large  as 
the  first,  what  are  the  numbers  ? 

5.  The  product  of  the  extremes  of  an  arithmetical  progression 
of  3  terms  is  4  less  than  the  square  of  the  mean.  What  are 
the  numbers,  if  their  sum  is  24  ? 

6.  The  sum  of  four  numbers  in  arithmetical  progression  is  14, 
and  the  product  of  the  means  is  12.     What  are  the  numbers  ? 

7.  The  sum  of  seven  numbers  in  arithmetical  progression  is 
98,  and  the  sum  of  their  squares  is  1484.    What  are  the  numbers  ? 


852  PROGRESSIONS 

8.  The  sum  of  five  numbers  in  arithmetical  progression  is  15, 
and  the  product  of  the  extremes  is  3  less  than  the  product  of  the 
terms  next  to  the  extremes.     What  are  the  numbers  ? 

9.  A  number  is  expressed  by  three  digits  in  arithmetical  pro- 
gression. If  the  number  is  divided  by  tlie  sum  of  its  digits,  the 
quotient  is  20^ ;  and  if  the  number  is  increased  by  594,  the  result 
is  the  number  with  its  digits  in  the  reverse  order.  What  is 
the  number? 

10.  Find  the  sum  of  the  odd  numbers  from  1  to  100. 

11.  The  product  of  the  extremes  of  an  arithmetical  progression 
of  10  terms  is  70,  and  the  sum  of  the  series  is  95.  What  are  the 
extremes  ? 

12.  Fifty-five  logs  are  to  be  piled  so  that  the  top  layer  shall 
consist  of  1  log,  the  next  layer  of  2  logs,  the  next  layer  of  3  logs, 
etc.     How  many  logs  must  be  placed  in  the  bottom  layer  ? 

13.  It  cost  Mr.  Smith  f  19.00  to  have  a  well  dug.  If  the  cost 
of  digging  was  $1.50  for  the  first  yard,  f  1.75  for  the  second, 
$  2.00  for  the  third,  etc.,  how  deep  was  the  well  ? 

14.  The  product  of  the  extremes  of  an  arithmetical  progression 
of  15  terms  is  93,  and  the  sum  of  the  first  and  last  means  is  34. 
What  is  the  progression  ? 

15.-  How  many  arithmetical  means  must  be  inserted  between 
5  and  37,  so  that  the  ratio  of  the  first  mean  to  the  last  mean 
may  be  ^j  ? 

16.  How  many  arithmetical  means  must  be  inserted  between 
4  and  25,  so  that  the  sum  of  the  series  may  be  116  ? 

17.  Prove  that  the  equimultiples  of  the  terms  of  an  arith- 
metical progression  are  in  arithmetical  progression. 

18.  Prove  that  the  difference  of  the  squares  of  consecutive 
integers  are  in  arithmetical  progression,  and  that  the  common 
difference  is  2. 

19.  Prove  that  the  sum  of  n  consecutive  odd  integers,  beginning 
with  1,  is  n\ 


PROGRESSIONS  353 


GEOMETRICAL  PROGRESSION 

366.  A  series  of  numbers  each  of  which  after  the  first  is  derived 
by  multiplying  the  preceding  number  by  some  constant  multiplier 
is  called  a  Geometrical  Series,  or  a  Geometrical  Progression. 

2,  4,  8,  16,  32  and  a*,  a^,  a^,  a  are  geometrical  progressions. 

In  the  first  series  the  constant  multiplier  is  2  ;  in  the  second  it  is  -• 

a 

G.  P.  is  an  abbreviation  of  the  words  Geometrical  Progression. 

367.  The  constant  multiplier  is  called  the  Ratio. 

It  is  evident  that  the  terms  of  a  geometrical  progression 
increase  or  decrease  numerically  according  as  the  ratio  is  numer- 
ically greater  or  less  than  1. 

368.  To  find  the  nth,  or  last  term. 

1.  In  the  geometrical  progression  3,  6,  12,  24,  what  is  the 
ratio  of  6  to  3  '/  of  12  to  6  ?  of  24  to  12  ? 

2.  In  the  geometrical  progression  a,  ar,  ar^,  ar',  •••  what  is  the 
ratio  ?  How  many  times  does  the  ratio  enter  as  a  factor  into  the 
second  term  ?  into  the  third  term  ?  into  the  fourth  term  ? 

369.  When  a  represents  the  first  term  of  a  G.  V.,  r  the  ratio, 
and  I  the  last  or  nth  term, 

I  =  ar^'-K  (I) 


Examples 
1.    Find  the  9th  term  of  the  series  1,  3,  9,  •••. 

PROCESS 

Explanation.  — In  this  example  a  =  1,  r  =  3,  and 

I  =zl  X  3^  Substituting  these  values  in  the  formula  for  I,  the 

,       r.,-ni  last  term  is  6561. 

( =  6561 

2.  Find  the  10th  term  of  the  series  1,  2,  4,  •••. 

3.  Find  the  8th  term  of  the  series  \,  ^,  1,  •••. 
'4.    Find  the  9th  term  of  the  series  6,  12,  24,  .... 

ALG,  —  23 


354  PROGRESSIONS 

5.  Find  the  11th  term  of  the  series  ^,  1,  2,  •••. 

6.  Find  the  7th  term  of  the  series  2,  6,  18,  •••. 

7.  Find  the  6th  term  of  the  series  4,  20,  100,  .... 

8.  Find  the  6th  term  of  the  series  6,  18,  54,  •••. 

9.  Find  the  10th  term  of  the  series  1,  ^,  \,  •••. 

10.  Find  the  10th  term  of  the  series  1,  |,  ^,  •••. 

11.  Find  the  8th  term  of  the  series  ^,  ^,  ^,  •••. 

12.  Find  the  11th  term  of  the  series  a^%,  o}%\  ■-. 

13.  Find  the  nth  term  of  the  series  2,  V2,  1,  •••. 

14.  A  man  worked  for  25  cents  the  first  day,  50  cents  the 
second  day,  f  1  the  third  day,  and  so  on  for  10  days.  How  much 
did  he  receive  the  tenth  day  ? 

15.  If  a  man  begins  business  with  a  capital  of  $200  and 
doubles  it  every  year  for  6  years,  how  much  will  he  have  at 
the  end  of  the  sixth  year? 

16.  If  the  population  of  the  United  States  is  76  millions  in 
1900  and  doubles  itself  every  25  years,  what  will  it  be  in  the 
year  2000? 

17.  A  man's  salary  was  raised  \  every  year  for  5  years.  If 
his  salary  was  $  512  the  first  year,  what  was  it  the  sixth  year  ? 

18.  The  population  of  a  city  at  a  certain  time  was  20,736,  and 
increased  in  geometrical  progression  25%  each  decade.  What 
was  the  population  at  the  end  of  40  years  ? 

19.  A  man  who  wanted  10  bushels  of  wheat  thought  $1  a 
bushel  too  high  a  price.  But  he  agreed  to  pay  2  cents  for  the 
first  bushel,  6  cents  for  the  second,  18  cents  for  the  third,  and 
so  on.     What  did  the  last  bushel  cost  him  ? 

20.  From  a  grain  of  corn  there  grew  a  stalk  that  produced 
an  ear  of  150  grains.  These  grains  were  planted,  and  each  pro- 
duced an  ear  of  150  grains.  This  process  was  repeated  until 
there  were  4  harvestings.  If  75  ears  of  corn  make  1  bushel,  how 
many  bushels  were  there  the  fourth  year  ? 


PROGRESSIONS  355 

370.   To  find  the  sum  of  a  finite  series. 

Let  a  represent  the  first  term,  I  the  nth.  term,  or  the  last  term, 
r  the  ratio,  n  the  number  of  terms,  and  s  the  sum  of  the  terms.  • 

Then,  s  =  a  +  ar  +  a/-^  +  a?-* -| \- ar'^'K  (1) 

(l)xr,  rs  —  ar  +  ar^  +  a7-^ -{- \- ar"~^ -\- ar\  (2) 

(2)-(l),      s(r-l)=ar"-a. 

(11) 


s  = 

r  - 

—  a 

a(?-" 
r  — 

-1). 
1 

But,  since  ar 

"-»  =  / 

ar 

"  =  rl. 

Substituting 

rl  for 

ar" 

in  (II), 

rl  —  a 

a 

-rl 

r-r 


or  ^^^-L'.  (Ill) 


Examples 

1.  Find  the  sum  of  6  terms  of  the  series  3,  9,  27,  •••. 

PROCESS 
f^iyM ^  Explanation.  ; —  Since  the  first  term  a,  the 

*  —     .  _  ^  ratio  r,  and  the  number  of  terms  n,  are  given, 

and  fonnula  II  gives  the  sum  in  terms  of  a,  r, 
^  _  o  X  o  —  o  _  -j^Qg2      ^.'^'i  ">  formula  II  is  used. 

2.  Find  the  sum  of  8  terms  of  the  series  1,  2,  4,  •••. 

3.  Find  the  sum  of  8  terms  of  the  series  1,  ^,  \,  •••. 

4.  Find  the  sum  of  10  terms  of  the  series  1,  1^,  2\,  •••. 

5.  Find  the  sum  of  7  terms  of  the  series  2,  —  |,  f,  •••. 

•  6.    Find  the  sum  of  12  terms  of  the  series  —  ^,  \,  ~  j,  •••. 

7.  Find  the  sum  of  7  terms  of  the  series  1,  2  a:,  4  a^,  •••. 

8.  Find  the  sum  of  7  terms  of  the  series  1,  —  2  a;,  4  a^,  •••. 

9.  Find  the  sum  of  n  terms  of  the  series  1,  a^,  x*,  •••. 
10.  Find  the  sum  of  n  terms  of  the  series  1,  2,  4,  •••. 


356  PROGRESSIONS 

11.  Find  the  sum  of  n  terms  of  the  series  1,  \,  \,  •••. 

12.  The  extremes  of  a  geometrical  series  are  1  and  729,  and  the 
ratio  is  3.     What  is  the  sum  of  the  series  ? 

13.  What  is  the  sum  of  the  series  3,  6,  12,  •••,  192  ? 

14.  What  is  the  sum  of  the  series  7,  -  14,  28,  •-,  -  224  ? 

371.    To  find  the  sum  of  an  infinite  geometrical  series. 

If  the  ratio  r  is  numerically  less  than  1,  it  is  evident  that  the 
successive  terms  of  a  geometrical  series  become  numerically  less 
and  less.  Hence,  in  an  iniinite  decreasing  geometrical  series,  the 
nth  term  I,  or  ar^~^,  can  be  made  less  than  any  assignable  number, 
though  not  absolutely  equal  to  zero. 

(Ill)  may  be  written  s  =  — —  ^ 


1-r 


Since,  by  taking  enough  terms,  I  and  consequently  rl  can  be 
made  less  than  any  assignable  number,  the  second  fraction  may 
be  neglected. 

Hence,  the  formula  for  the  sum  of  an  infinite  decreasing  geomet- 
»:ical  series  is 

a 

Examples 

1.  Find  the  sum  of  the  series  1,  y^^,  y^^,  •••. 

Solution 
Substituting  1  for  a  and  ^^  for  r  in  (IV) , 

s  =  — 1— =  — =  — 

2.  Find  the  value  of  .185185185  .... 

Solution 

Since  .185185185  •••  =  .185  +  .000185  +  .000000185  +  ...,  a  =  .185  and 
J- =  .001. 

Substituting  in  (I V) ,  .  185186185  • . .  =  s  =  '^^'^  =  — • 
^  ^  1  -  .001  27 


PROGRESSIONS  357 

Find  the  value  of 

3.  1  +  1  +  1+....  6.  .407407.... 

4.  3  +  f  +  y3^+....  7.  .363636. ... 

5.  i_^_i_| .  8.  1.94444  =^ 

372.  To  insert  geometrical  means  between  two  terms. 

Examples       *    •' 

ft 

1.  Insert  3  geometrical  means  between  2  and  162. 

PROCESS  Explanation.  —  Since  there  are  three  means,  there  are 

;  --  (ifn-\  five   terms,  and   n  —  1  =  4.     Solving  for  r  and   neglecting 

1  fio  _  9  ^  imaginary  values,  r  =  ±  3. 

IbZ  —  Z  r*  Therefore,  the  series  is  either  2,  6,  18,  54,  162,  or  2,  -6, 

r  =  ±  3  18,-54,  162. 

2.  Insert  3  geometrical  means  between  1  and  625. 

3.  Insert  5  geometrical  means  between  4^  and  -fy^. 

4.  Insert  4  geometrical  means  between  ^^  and  |-|. 

5.  Insert  4  geometrical  means  between  5120  and  5. 

6.  Insert  4  geometrical  means  between  4V2  and  1, 

7.  Insert  5  geometrical  means  between  r/  and  h^. 

8.  Insert  6  geometrical  means  between  —  2  and  ^  V2. 

9.  Insert  4  geometrical  means  between  x  and  —  y. 

373.  Principle.  —  The  geometrical  mean  between  two  numbers 
is  equal  to  the  square  root  of  their  product. 

The  above  principle  may  be  established  as  follows : 

Let  a  and  b  represent  any  two  numbers,  and  G  their  geometrical  mean. 
It  is  to  be  proved  that  G  =  y/ab. 

Since  the  two  numbers  and  their  geometrical  mean  form  the  geometrical 
progression  a,  G,  b, 

§  366.  ^  =  A, 

^  ^  a      G' 

'♦     f       '•  G^^ab. 

.:  G  =  \/ab. 


V'-N 


358  PROGRESSTOIVS 

Find  the  geometrical  meaxi  between 

1.  8  and  50.  4.    (a  +  bf  and  (a  -  6)*. 

2.  4^  and  34.  ox  ,  ,? 
2  ^  ^  a^  +  ab  .  ab  +  b^ 
^  , ,        1  o  5.    — — ' and — -• 

3.  11^  and  f.                              a'-ab  ab-b^ 

6.    25  a^  -  10  a;  +  1  and  x-  +  10  a-  +  25. 

374.  Since  formulae  I  and  II,  or  III,  which  is  equivalent  to 
II,  are  two  independent  simultaneous  equati)|ns  containing  five 
elements,  if  three  elements  are  known,  the  other  two  may  be  found 
by  elimination. 

Problems 
376.    1.    Given  r,  I,  and  s,  to  find  a. 

2.  The  ratio  of  a  geometrical  progression  is  5,  the  last  term  is 
625,  and  the  sum  is  775.     What  is  the  first  term  ? 

3.  The  ratio  of  a  geometrical  progression  is  ^,  the  sum  is  \, 
and  the  series  is  infinite.     What  is  the  first  term  ? 

4.  Find  I  in  terms  of  a,  r,  and  s. 

5.  Find  the  last  term  of  the  series  5,  10,  20,  •••,  the  sum  of 
whose  terms  is  155. 

6.  If  1+  1^  +  ^-1 =  1|.  (1  +  y^),  what  is  the  last  term, 

,  and  the  number  of  terms  ? 

7.  Deduce  the  formula  for  r  in  terms  of  a,  I,  and  s. 

8.  If  the  sum  of  the  geometrical  progression  32  •••  243  is  665, 
what  is  the  ratio  ?     Write  the  series. 

9.  The  sum  of  a  geometrical  progression  is  700  greater  than 
the  first  term  and  525  greater  than  the  last  term.  What  is  the 
ratio ?     If  the  first  term  is  81,  what  is  the  progression? 

10.  Deduce  the  formula  for  r  in  terms  of  a,  n,  and  I. 

11.  The  first  term  of  a  geometrical  progression  is  3,  the  last 
term  is  729,  and  the  number  of  terms  is  6.  What  is  the  ratio  ? 
Write  the  series. 

12.  Find  I  in  terms  of  r,  n,  and  s. 


PROGRESSIONS  359 

13.  The  sum  of  the  12  terms  of  a  geometrical  progression  whose 
ratio  is  2  is  4095.     What  is  the  12th  term  ? 

14.  The  velocity  of  a  sled  at  the  bottom  of  a  hill  is  100  feet 
per  second.  How  far  will  it  go  on  the  level,  if  its  velocity 
decreases  each  second  \  of  that  of  the  previous  second  ? 

15.  From  a  cask  of  vinegar  ^  was  drawn  off  and  the  cask  was 
filled  by  pouring  in  water.  Show  that  if  this  is  done  6  times, 
the  contents  of  the  cask  will  be  more  than  y^^  water. 

16.  A  ball  thrown  vertically  into  the  air  100  feet  falls  and 
rebounds  40  feet  the  first  time,  16  feet  the  second  time,  and  so 
on.  What  is  the  whole  distance  through  which  the  ball  will 
have  passed  when  it  finally  comes  to  rest  ? 

17.  Show  that  the  amount  of  $  1  for  1,  2,  3,  4,  5  years  at  com- 
pound interest  varies  in  geometrical  progression. 

18.  Show  that  equimultiples  of  numbers  in  geometrical  pro- 
gression are  also  in  geometrical  progression. 

19.  The  sum  of  three  numbers  in  geometrical  progression  is 
19,  and  the  sum  of  their  squares  is  133.     WTiat  are  the  numbers  ? 

Suggestion.  —  When  there  are  but  three  terms  in  the  series  tliey  may  be 
represented  by  x^,  xy,  y^,  or  by  x,  Vary,  y. 

20.  The  product  of  three  numbers  in  geometrical  progression 
is  8,  and  the  sum  of  their  squares  is  21.  What  are  the  three 
numbers  ? 

21.  If  4  is  a  geometrical  mean  between  two  numbers  whose 
sum  is  10,  what  are  the  numbers  ? 

22.  The  product  of  three  numbers  in  geometrical  progression 
is  64,  and  the  sum  of  their  cubes  is  584.    What  are  the  numbers  ? 

23.  The  sum  of  the  first  and  second  of  four  numbers  in  geo- 
metrical progression  is  15,  and  th&  sum  of  the  third  and  fourth 
is  60.     What  are  the  numbers  ? 

Suggestion.  —  Four  unknown  numbers  in  geometrical  progression  may 

a;2  w2 

be  represented  by  — ,  x,  y,  —  • 
y  25 

24.  The  sum  of  the  first  and  third  of  three  numbers  in  geo- 
metrical progression  is  130,  and  their  product  is  625.  What  are 
the  numbers  ? 


360  PROGRESSIONS 

25.  Divide  $700  among  three  persons  so  that  the  first  shall 
receive  $  300  more  than  the  third,  and  the  share  of  the  second 
shall  be  a  geometrical  mean  between  the  shares  of  the  first  and 
third. 

26.  If  a,  h,  and  c  are  in  geometrical  progression,  show  that 
their  reciprocals  also  are  in  geometrical  progression. 

27.  The  difference  between  two  numbers  is  24,  and  their 
arithmetical  mean  exceeds  their  geometrical  mean  by  6.  "What 
are  the  numbers  ? 

HARMONICAL  PROGRESSION 

376.  1.  Examine  the  series  1,  \,  \,  \,  •••.  Has  it  a  constant 
difference  ?     Has  it  a  constant  ratio  ? 

2.  Take  the  reciprocal  of  each  term.  What  kind  of  a  series 
is  thus  formed?  How,  then,  may  the  series  1,  \,  ^,  -^,  •••,  be 
described  ? 

377.  A  series  the  reciprocals  of  whose  terms  form  an  arith- 
metical progression  is  called  a  Harmonical  Series,  or  a  Harmonical 
Progression. 

3,  \i  1,  f,  f)  h  ■••  is  a  harmonical  progression,  because  \,  f,  1,  f,  |,  2,  ••■ 
*he  reciprocals  of  its  terms  form  an  arithmetical  progression. 

H.  P.  is  an  abbreviation  for  the  words  Harmonical  Progression. 

378.  Problems  in  harmonical  progression  are  commonly  solved 
by  taking  the  reciprocals  of  the  terms  and  employing  the  prin- 
ciples of  arithmetical  progression.  There  is  no  general  method, 
however,  for  finding  the  sum  of  the  terms  of  a  harmonical  pro- 
gression. 

379.  Principle  1.  —  The  harmonical  mean  between  two  numbers 
is  equal  to  twice  their  product  divided  by  their  sum. 

The  above  principle  may  be  established  as  follows : 
£et  H  represent  the  harmonical  mean  between  a  and  6. 
It  is  to  be  proved  that  H=     ^ 


a  +  6 


§  377,  -,   — ,   -  are  in  arithmetical  progression. 

a    H    b 


PROGRESSIONS  361 


Hence,  §  356, 

1       1  _  1  _1. 
b     H     H     a 

Clearing  of  fractions, 

aH-ab^ah-  bH. 

Transposing, 

aH+bH=2ab. 

.-.  if_2«^ 

§364, 

^-     2 

§373, 

G'=\/^. 

§  379, 

a +  6* 

Multiplying  (1)  by  (3), 

AH^ah. 

Taking  the  square  root, 

yjAH=\'ab. 

From  (2)  and  (5),  Ax.  1, 

a  =  y/AH. 

a  +  h 

380.  Principle  2.  —  The  geometrical  mean  between  two  numbers 
is  also  the  geometrical  mean  between  their  arithmetical  and  harmoni- 
cal  means. 

The  above  principle  may  be  established  as  follows : 

(1) 

(2) 
(3) 

(4) 
(6) 


Hence,  §  373,  G  is  the  geometrical  mean  between  A  and  H. 

Examples 

1.  Find  the  12th  term  of  the  H.  P.  6,  3,  2,  .... 

Solution.  — The  reciprocals  of  the  terms  form  the  arithmetical  progression 

f)  5>  2»      • 
in  which  a  =  J  and  d  =  |. 

Substituting  ^  for  a,  J  for  d,  and  12  for  n  in  (I) , 

§359,  Z  =  ^  +  (12 -1)1  =  2. 

Therefore,  §  377,  the  12th  terra  of  the  given  harmonical  progression  is  J. 

2.  Find  the  10th  term  of  the  harmonical  series  ^,  \,  ^,  •••. 

3.  Insert  6  harmonical  means  between  1^  and  12. 

4.  Insert  2  harmonical  means  between  2  and  5. 

5.  Insert  7  harmonical  means  between  12^  and  2^. 


362     •  PROGRESSIONS 

6.  Insert  3  hannonical  means  between  b  and  a. 

7.  Find  the  nth  term  of  the  H.  P.  |,  i,  j\,  .... 

8.  The  3d  and  4th  terms  of  a  H.  P.  are  2i  and  1|-,     Write  the 
first  6  terms. 

Find  the  harraonical  mean  between 

9.  2  and  3.  13.    a  —  c  and  a  +  c. 

10.  I  and  I.  14.    1  —  Va  and  1  +  \^. 

11.  21  and  1^.  15.    o  and  -. 

a 

12.  21  and  10.  ^g     V6  and  V3. 

17.  The  5th  term  of  a  harmonical  progression  is  -^j,  and  the 
]  1th  term  is  2^-     What  is  the  first  term  ? 

18.  The   arithmetical   mean   between  two  numbers  is  5,  and 
their  harmonical  mean  is  3^.     What  are  the  numbers  ? 

19.  If  one  number  exceeds  another  by  2,  and  their  arithmetical 
mean  exceeds  their  harmonical  mean  by  ^,  what  are  the  numbers  ? 

20.  If  a,  b,  and  c  are  in  harmonical  progression,  prove  that 
a  —  b:b  —  c  =  a:c. 

21.  If  a,  &,  c,  and  d  are  in  harmonical  progression,  prove  that 
ab:  cd  =  b  —  a:d~c. 

22.  If  6  is  the  harmonical  mean  between  a  and  c,  prove  that 

1  111 

+  i -=-  + 


6  —  a     b  —  c     a     c 

23.  When  b  —  a:c  —  b  =  a:x,  prove  that  x  =  a,  if  a,  b,  and  c 
are  in  arithmetical  progression ;  that  x  =  b,  if  a,  b,  and  c  are  in 
geometrical  progression;  and  that  x  —  c,  if  a,  b,  and  c  are  in 
harmonical  progression. 

24.  The  harmonical  mean  between  two  numbers  is  5^  and  their 
arithmetical  mean  is  6^.     What  is  their  geometrical  mean  ? 

25.  Prove  that  x  +  xy,  2xy,  and  xy -\- xy^  are  in  harmonical 
progression. 

26.  If  b  -\-  c,  c  +  a,  and  a  -\-b  are  in  harmonical  progression, 
prove  that  a?,  b'\  c-  are  in  arithmetical  progression. 


IMAGINARY   AND    COMPLEX    NUMBERS 


381.  1.  If  from  V—  25  the  rational  factor  V25  is  removed, 
what  irrational  factor  remains  ? 

2.  Simplify  V— 25,  V— 16,  V—  a^.  What  common  part,  or 
unit,  have  the  indicated  square  roots  of  negative  numbers  ? 

3.  What  is  the  square  of  Vi?  of  V5?  of  V9?  of  y/2x? 
What  is  the  effect  of  squaring  a  radical  of  the  second  degree  ? 

What,  then,  is  the  square  of  V—  4 ?   of  V—  a?   of  V—  1  ? 

382.  Up  to  this  point  the  only  numbers  whose  nature  has  been 
discussed  have  been  numbers  that  differ  from  arithmetical  num- 
bers in  having  a  sign,  +  or  — ,  to  indicate  quality  or  direction. 
These  numbers  are  called  real  numbers,  and  may  be  briefly 
described  as  numbers  ivhose  squares  are  x>ositive. 

There  are  numbers,  however,  whose  squares  are  negative.  They 
constitute  one  class  of  imaginary  numbers,  defined  in  §  257.  In 
this  chapter  only  imaginary  numbers  of  the  second  degree  are 
treated. 

Let  —  a  be  any  negative  real  number. 

Then,  V—  a  will  represent  an  imaginary  number. 

Since         +  V—  a  =  +  V«  •  V—  1  =  +  V—  1  •  Va 

and  —  V—  a  =  —  ^a  •  V—  1  =  — V—  1  •  Va, 

the  positive  imaginary  unit  is  +  V—  1,  and  the  negative  imagi- 
nary unit  is  —V—  1. 

Since  the  square  of  the  square  root  of  a  number  is  the  number 
itself, 

This  relation  is  sufficient  to  explain  operations  with  imaginary 
numbers. 

363 


-x  ^n 


'^ 


364  IMAGINARY  AND   COMPLEX   NUMBERS 

383.    Relation  between  the  units  +1,  —  1,  V—  1,  and  —  V—  1. 

In  the  accompanying  figure  the  length  of  any  radius  of  the  circle  represents 
the  arithmetical  unit  1.     The  line  drawn  from  0  to  A,  called  the  line  OA, 
^  represents  the  positive  unit  + 1,  and 

the  line  OA"  represents  the  nega- 
tive unit  —  1.  Every  real  number 
lies  somewhere  on  the  line  X'X, 
which  is  supposed  to  extend  indefi- 
nitely in  both  directions  from  0. 

' -Y    X'Xis  called  the  axis  of  real  num- 

bers. 

The  direction  of  any  line  drawn 
from  0,  as  OB,  that  is,  the  quality 
or  direction   sign   of   the   number 
represented  by  that  line,  is  deter- 
^  mined  with  reference  to  the  fixed 

line  OA  by  finding  what  part  of  a  revolution  is  required  to  swing  the  line 
from  the  position  OA  to  the  required  position.  By  common  consent  revo- 
lution of  the  line  OA  is  performed  in  a  direction  opposite  to  that  of  the  hands 
of  a  clock,  as  shown  by  the  arrows.  OB  is  reached  after  ^  of  a  revolution, 
OA'  after  \  of  a  revolution,  OA"  after  J  of  a  revolution,  etc. 

Since  OA",  or  —1,  represents  ^  of  a  revolution  of  OA,  the  square  of  OA", 
or  (—  1)2,  represents  1  revolution  of  OA,  which  produces  OA,  or  +  1. 
Hence,  OA",  or  —  1,  represents  the  square  root  of  +  1,  or  (+  1)^. 

Similarly,  since  OA'  represents  J  of  ^  of  a  revolution  of  OA,  and  OA" 
represents  ^  of  a  revolution  of  OA,  OA'  represents  the  square  root  of  OA", 
or  of  -  1  ;  that  is,  OA'  =  V^. 

If  OA"  is  swung  J  of  a  revolution  from  the  position  OA"  to  the  position 
OA'",  OA"  will  be  multiplied  by  V—  1  just  as  OA  is  multiplied  by  V—  1  to 
produce  OA'.     Hence,  the  result  OA'"  =  —  1  •  V—  1  =  —  V—  1. 

+  1,  represented  by  OA,  and  —  1,  represented  by  OA",  are  the  units  for 
real  numbers,  that  is,  are  real  units.  Just  as  the  real  number  +  a  is  repre- 
sented by  a  line  a  units  long  extending  from  0  toward  X,  and  the  real  num- 
ber —  a  by  a  line  a  units  long  extending  from  O  in  the  opposite  direction,  so 
the  imaginary  number  +  av^^l,  or  (-|-V—  1)  x  a,  is  represented  by  a  line 
a  units  long  extending  from  0  toward  Y,  and  the  imaginary  number  —  a  V  —  1 , 
or  (  — V—  1)  X  a,  by  a  line  a  units  long  extending  from  0  in  the  opposite 
direction,  toward  Y'.  Hence,  +V—  1  and  —  V—  1  are  the  units  for  imagi- 
nary numbers,  that  is,  they  are,  imaginary  units ;  +  aV—  1  is  called  a  posi- 
tive imaginary  number  and  —  a  V—  1  a  negative  imaginary  number. 

Y'  Y  is  called  the  axis  of  imaginary  numbers.  If  Y'  Y  is  taken  as  the 
axis  of  real  numbers,  then  X'X  becomes  the  axis  of  imaginary  numbers. 
Hence,  it  is  seen  that  imaginary  numbers  have  as  much  reality  as  real  num- 
bers.    Imaginary  uumbers  were  named  before  their  nature  was  understood. 


IMAGINARY  AND   COMPLEX  NUMBERS 


365 


384.  In  the  graphical  illustration  of  the  relation  between  real 
and  imaginary  numbers  it  was  assumed  that  -f  1  •  V—  1  =  V— 1, 
or  V—  1  •  1 ;  that  is,  the  Commutative  Law  for  multiplication  was 
assumed  to  apply  when  imaginary  numbers  were  involved.  It  is 
evident,  if  the  discussion  of  number  is  to  proceed,  that  in  any 
operation  imaginary  numbeijs  must  obey  all  the  laws  of  real  num- 
bers except  those  which  determine  the  quality  of  the  result ;  and 
that  the  quality  of  the  result  is  determined,  as  far  as  the  imaginary 
numbers  are  concerned,  by  the  relation  (V—  1)*'  =  —  1. 

385.  Powers  of  V^^. 

(V^)^  =-1; 

and  so  on.     Hence,  if  n  =  0  or  a  positive  integer, 

( V^rT)*n+l  ^  ^  V"3T  ;      ( V^T)^"+2  =  - 1 ; 

(V^^)*"+^  =  -V^^;    ( V^^)*"+*  =  + 1. 

Hence,  any  even  ])oiver  o/  V—  1  is  real  and  any  odd  power  is 
imoffinary. 

386.  Operations  involving  imaginary  numbers. 


Examples 
Find  the  value  of 

2.    (V^^)^        4.    {-y/^^y\        6.    (V^^)^'.        8.    (-V^^ 


9.    Add  V— a*  and  V— 16  a*. 

Solution 

V-a*  +  V-  16  a*  =  a^y/^^  +  4  aV^  =  5  aV^^. 


366  IMAGINARY  AND   COMPLEX  NUMBERS 

Simplify  the  following : 


10.  V^4-f-V-49.  13.    V^n^  +  4V^^. 

11.  V^l)+V-64.  14.    5V^18-V^72. 

12.  2V^^  +  3V^^.  15.    3V^^-V^80. 

16.  ( V'^^  +  SV^b)  +  ( V^^  -  3  V^^). 

17.  (V—  9a;y  —  V—  xy)  —  (V— 4a;?/ +  V— x?/). 


18.    V-a^  +  V-4ic2-V-ar*  +  3a;V- 


X. 


19.    V-  16  -  3V^^  +  V-  18  +  V-  50  +  V-  25. 


20.  V^8  +  aV^2-V-98-5V^=='2^^. 

21.  V-  16  aV  +  V-  aV  _  ^_  g  ^2^ 


22.    V1-5-3V1-10  +  2V5-30. 


23.    Multiply  3V-10  by  2V- 3. 

PROCESS 

3 V^IO  X  2V^^  =  3 VlO V^^  X  2V8V^^ 


=  6V10  X  8  X  (-  1) 

=  -  6  V80  =  -  24  V5. 

ExPLANATiox.  —  In  order  to  determine  the  sign  of  the  product,  each  imagi- 
nary number  is  reduced  to  the  form  bV  —  1.  The  numbers  are  then  multiplied 
together  as  ordinary  radicals,  observing,  however,  that  V—  1  x  V—  1  =  —  1. 

24.  Multiply  V^2  +  3 V^^  by  4 V^^  -  V^3. 

First  Solution  Second  Solution 

(V2  +  3V3)(4V2-\/3)(-l)  -4V4-  12V6        « 

=  (8  +  12V6->/6-9)(-l)  +3V9+       V6 

=   1-11^^  ~l        _  IIV6 

Multiply : 

25.  3V^5  by  2V^15.  28.    SV^l  by  V^^. 

26.  4V-  27-  by  V-12.  29.    V^l25  by  V^^^IOS. 

27.  2V^^  by  5V^^.  30.    V^^^^  by  V^^^30. 


IMAGINARY  AND   COMPLEX  NUMBERS 


367 


31.  v^::6+V^^  by  V^6-V^^. 

32.  V— o6  +  V  —  a  by  V—  ab  —^—a. 

33.  V  —xy  +  V—  a;  by  V—  xy  +  V—  a;. 

34.  V^r^-V^=T2by  V^^-V"^75. 

35.  V— a +V— 6 +V— c  by  V—  a  +  V—  6  —  V— c. 

36.  Divide    V— 12  by  V^^. 

Solution 


37.   Divide   Vl2  by 


V3 


:  =  V4  =  2. 


Solution 

\/l2  V4 


V-3     VsV^l 


38.   Divide  5  by  (V-  1/. 


V3T     v^T 


1 


Solution 
5        _  5(+  1)   _  5(V^^)^ 

(>/^=n:)8    (v^8    (>/^)3 


2\/-l. 


dV^n.. 


Divide : 

39.  V-18  by  V^^. 

40.  V27  by  V^^. 

41.  14V^^  by  2V^7. 

42.  —  V-a-  by  V-  6^. 

43.  1  by  V^n!. 

44.  (V^)*-V^  by  V^. 

45.  V^+(V^)'  by  V^. 

46.  V8  -  3  Vl4  by  V^^. 

47.  Vi2+V3  by  V^.   • 

57. 


48.  -  2  by  V^^. 

49.  (V^^)'  by  iV^=n:. 

50.  (V^^y  by  (V^^^ 

51.  V4a6  by  V— 6c. 

52.  V-20- V"^  by  2V^. 
53. 
54. 


V^16-V-6  by  2V-2. 
(V3T)H  by   _  i^ZIl, 

55.  (V^^^)'"  by  (V^^)-l 

56.  V-a'+bV^l  by  V— a6. 
4  by  V^2  .  V^^  •  V^^. 


368  IMAGINARY  AND   COMPLEX  NUMBERS 

387.  For  brevity  V  —  1  is  often  written  i. 

Including  all  intermediate  fractional  and  incommensurable 
values,  the  scale  of  real  numbers  may  be  written 

.■.._3 2 1...0...  +  1...  +  2..-  +  3.-.  (1) 

and  the  scale  of  imaginary  numbers,  composed  of  real  multiples 
of  +  i  and  —  i,  may  be  written 

Zi 2  i i  ...0 ...  +  i--'  +  2i ...  +  3/..-  (2) 

Since  the  square  of  every  real  number  except  0  is  positive  and 
the  square  of  every  imaginary  number  except  0  i,  or  0,  is  negative, 
the  scales  (1)  and  (2)  have  no  number  in  common  except  0.    Hence, 

An  imaginary  number  cannot  he  equal  to  a  real  number  nor  cancel 
any  part  of  a  real  number. 

388.  The  algebraic  sum  of  a  real  number  and  an  imaginary 
number  is  called  a  Complex  Number. 

2  +  3  V  —  1,  or  2  +  3  i,  and  a  +  h  V  —  1,  or  a  +  hi,  are  complex  numbers. 
a^  +  2aby/  —  \  —b^  is  a  complex  number,  since  a^  +  2  a6  V  —  1  —  62  — 
(a2  -  62)  +  2  ah  y/"^^. 

389.  Two  complex  numbers  that  differ  only  in  the  signs  of 
their  imaginary  terms  are  called  Conjugate  Complex  Numbers. 

a  +  6  V  —  1  and  a  —  6  V  —  1,  or  a  -\-  hi  and  a  —  6i,  are  conjugate  com- 
plex numbers. 

390.  Tlie  sum  and  product  of  two  conjugate  complex  numbers  are 
both  real. 

Let  a  +  hV  —  1  and  a  —  b  V—  1  be  conjugate  complex  numbers. 

Their  sum  is  2  a. 

Since  (V  —  1)2  =  —  1,  their  product  is, 

§97,  c2-(6V^^)2  =  a2-(-62) 

=  a-  +  62. 

391.  If  two  complex  numbers  are  equal,  their  real  parts  are  equal 
and  also  their  imaginary  parts. 

Let  a  +  b  V^^  =  x  +  y  V-  1. 

Then,  a  -  x  =  (y  -  b)  V^, 

which,    §  387,  is  impossible  unless  a  =  x  and  y  =  b. 


IMAGINARY  AND   COMPLEX  NUMBERS 


369 


392.  //"a  +  ftV  —  1  =  0,  a  and  h  being  real,  then  a  =  0  and  6  =  0. 

For,  squaring,  a^^  4-  2  o6  V  —  1  —  6^  =  o, 

a2  _  52  ^  _  2  a6  V^H, 
which,  §  387,  is  true  only  when  a  =  0  and  6  =  0. 

393.  Graphical  representation  of  a  complex  number. 

The  sum  of  3  positive  real  units  and  2  positive  imaginary  units  is  found  by 
counting  3  units  along  OX  in  the  positive  direction  from  0  and  from  that 
point,  Z),  measuring  2  units  upward  at 
right  angles  to  OX  in  the  direction  of 
the  axis  of  imaginary  numbers.  The 
line  OP  represents,  by  its  length  and 
direction,  the  combined  effect  or  sum 
of  the  directed  lines  OD  and  DP,  that 
is,  the  complex  number  3  +  2 1. 

The  same  result  may  be  obtained 
by  counting  2  units  along  OY  up- 
ward from  O  and  from  the  end  of 
the  second  division  measuring  3  units 
toward  the  right  at  right  angles  to 
0  Y  in  the  direction  of  the  axis  of  real  numbers, 
sents  either  3  +  2  i  or  2  i  +  3. 

Similarly,  the  line  OP'  represents  by  its  length  and  direction  2\  —  \i  or 
—  J  I -f  2 J,  and  the  line  OP'  represents  —  J  +  i  or  i  —  J. 

Represent  the  following  numbers  graphically : 

1.  3 +  4 1.  3.   5  +  2i.  5.   1-2. 

2.  2-3?".  4.   b-2i.  6.    4i-l. 


Hence,  the  line  OP  repre- 


394.    Relation  of  complex  numbers  to  real  and  imaginary  numbers. 

Let  a  and.  6  represent  any  real  numbers. 

In  the  figure  of  §  393  let  P  represent  any  point  a  units  dis- 
tant from  Y'  Y  and  h  units  distant  from  X'X. 

Then  OP,  or  the  complex  number  a  +  h  V  —  1,  represents  any 
number  whatever. 

If  P  lies  on  the  axis  of  real  numbers,  6  =  0  and  the  complex 
number  a  -f-  6  V  —  1  =  a,  a  real  number. 

If  P  lies  on  the  axis  of  imaginary  numbers,  a  =  0  and  the  com- 
plex number  a  +  6  V  —  1  =  6  V  —  1,  an  imaginary  number. 

If  P  lies  in  both  axes,  a  =  0  and  6  =  0,  and  the  complex  num- 
ber a  +  h  V—  1  =  0. 

ALG.  — 24 


370 


IMAGINARY  AND   COMPLEX  NUMBERS 


395.   Operations  involving  complex  numbers. 

Examples 
1.    Add  3  -  2V^i:  and  2  +  SV^T. 

Solution 

3  -  2\/:r"i  +  2  +  5V^^  =  (3  +  2)  +  (- 2\/^^  4- 5 V^n:) 

=  5  +  3V^. 

ExPLANATiox.  —  Since,  §  387,  the  imaginary  terms  cannot  unite  with  the 
real  terms,  the  simplest  form  of  the  sum  is  obtained  by  uniting  the  real  and 
the  imaginary  terms  separately  and  indicating  the  algebraic  sum  of  the  results. 


Simplify 

the  following : 

2. 

(5+V= 

■4)  +  (V-9- 

3). 

3. 

(2-V= 

.16)  +  (3+V- 

-4). 

4. 

(3-V^ 

r8)  +  (4+V^ 

-  18). 

5. 

(V-20 

-V16)  +  (V^ 

-45+V4). 

6. 

(4+V= 

•25)-(2+V- 

-4). 

7. 

(3-2V 

_5)-(2-3V-5). 

8. 

(2-2V 

-l+3)-(Vl6-V-16), 

9. 

V-49- 

_2-3V-4 

-V-1  +  6. 

10.    Expand  (a  +  &V—  l)(a  +  &V—  1). 

Solution 
§  91,  (a  +  6\/^n)(a+  &v'^=n')=  a^  +  2  aftV^H  4.(6v'3i)2. 

§384,  =a2  +  2a6V=^- 62. 


11.    Expand  (Vo  —  V-  3)^ 


Solution 


(V5 


3)2  =  5_2V^rT5  4.(_3) 
=  (5_3)-2V-  16 
=  2-2V-15. 


Expand  the  following : 

12.    (2  +  3V^T)(l+V^^). 

13.  (5  _  v^n;)(i  -  2V^i). 

14.  (V2+V^:^)(V8-V^^). 


15.  {2  +  Zif. 

16.  (2-3i/. 

17.  (a  -  Uf. 


IMAGINARY  AND   COMPLEX  NUMBERS 


371 


18.  Show  that  (1  -f.V^^)(l  +V^^)(1  +V^^)=-8. 

19.  Show  that  (-  1  +  V^=^)(-  1  +V^^)(-  1  -I- V^^) 

20.  Show  that  (-  i  +  iV^(-Hi V^3)(-|+i-V^) 

21.  Divide  8  +  V^^  by  3  +  2V^^. 

First  Solution 


=  1. 


3+2\/-l 


2-V-l 


8+aA^  =  6+    V-1  +  2 
6  +  4^^^! 

-3\/^n:  +  2 

-  SV^T  +  2 


The  real  term  of  the  dividend  may  always  be  separated  into  two  parts, 
of  which  will  exactly  contain  the  real  term  of  the  divisor. 


Second  Solution 
3  +  2-^/^:^1      (3  +  2V":n)(3-2\/^n;)  9  +  4 


=  2-^-1. 


26.  16  +  4  V^r5  by  3  -  V - 

27.  a^  +  6^  by  a  -  feV^^. 

28.  a  +  hi  by  ai  +  h. 

29.  (1  +  0'  by  1  -  i. 


Divide : 
^22.   3  by  1  -V^^. 

23.  2  by  1  +V-  1. 

24.  4+V4by2-V^^. 

25.  9  +  V^^  by  3  -  V- 
30.   Find  by  inspection  the  square  root  of  3  +  2V—  10. 

Solution 
3  +  2V:n[0=(5-2)+2V5-  -2  =  5  +  2V5.  -2 +(-2). 

.-.  V3  +  2>/^^=V5  +  2V5.  -2+(-2)  =  V5+  V^. 
Find  by  inspection  the  square  root  of 
/31.   4  +  2V^=^2T.     33.    6-2V^=^.       35.    12V^n:-5. 
32.    1+2V^6.       34.    9-f  2V^^2.  /36.    h''  +  2ahV^^- 

yi.   Verify  that  —  1  +  V—  1  and  —  1  — V—  1  are  roots  of 
equation  a^  +  2ic  +  2  =  0. 

38.   Expand  (|  +  ^V^r3)3. 


the 


372.  IMAGINARY  AND   COMPLEX  NUMBERS 

396.    If  Va  -\-bi  =  \/x  +  iy/y,  then  Va  —  hi  =  y/x  —  i^/y,  when 
a  and  b  are  real  and  x  and  y  are  positive. 

For  suppose  that  Va  +  hi  =  Vx  +  iVy. 

Squaring,  a  +  hi  =  3?  +  2  iy/xy  —^ 

.'.  §  391,  a  =  35  —  y  ana~Br=  2  i\/xy. 

a  —  bi  =:x  —  2  iVxy  — 
=  (Vx  —  iVyy. 
Hence,  Va  —  bi  =  Vx  —  i  Vy. 

Examples 
1.   Find  the  square  root  of  —  5  —  12 V—  1. 

Solution 


Let  Vx-iVy  =  V- 5- 12V^^.  (1) 

Then,  §  396,  Vx  +  iVy  -  V-  5  +  12  V^^.  (2) 

Multiplying,  x  -\-  y  =  V25  +  144, 

or  X  +  2/  =  13.  (3) 

Squaring  (1),  x  —  2  iy/xy  —  ?/  =  —  5  —  12  V—  1. 

Therefore,  §  391,  a;  -  ?/  =  -  5,  (4) 

Solving  (3)  and  (4),  x  =  4,  y  =  9. 

.-.   Vx  =  2,  iy/y  =  3  V—  1. 
Hence,  by  (1),        V- 5  -  12V^  =  2  -  3V^. 
This  result  may  be  verified  by  squaring  2  —  3  V—  1. 

2.    Find  the  square  root  of  —  5  +  12V—  1. 

Solution 
By  the  preceding  example,   v  —  5  —  12  V— 1  =2—  3V—  1. 


Therefore,  §  396,  V-5  +  l^yT^  =  2  +  3 V^^. 

Find  the  square  root  of 


INEQUALITIES 


397.  One  number  is  said  to  be  greater  than  another  when  the 
remainder  obtained  by  subtracting  the  second  from  the  first  is 
positive,  and  to  be  less  than  another  when  the  remainder  obtained 
by  subtracting  the  second  from  the  first  is  negative. 

If  a  —  ft  is  a  positive  number,  a  is  greater  than  6  ;  but  if  a  —  ft  is  a  negative 
number,  a  is  less  than  ft. 

Any  negative  number  is  regarded  as  less  than  0 ;  and,  of  two 
negative  numbers,  that  more  remote  from  0  is  the  lessi "' 

Thus,  —  1  is  less  than  0 ;   —  2  is  less  than  —  1  ;  —  3  is  less  than  —  2  ;  etc. 

An  algebraic  expression  indicating  that  one  number  is  greater 
or  less  than  another  is  called  an  Inequality. 

398.  The  Sign  of  Inequality  is  >  or  <. 

It  IS"  placed  between  two  unequal  numbers  with  the  opening 
toward  the  greater. 

Thus,  a  is  greater  than  ft  is  written  «  >  ft  ;  a  is  less  than  ft  is  written  a<,b. 

The  expressions  on  the  left  and  right,  respectively,  of  the  sign 
of  inequality  are  termed  the  Jirst  and  second  members  of  the 
inequality. 

399.  When  the  first  members  of  two  inequalities  are  each 
greater  or  each  less  than  the  corresponding  second  members,  the 
inequalities  are  said  to  subsist  in  the  same  sense. 

When  the  first  member  is  greater  in  one  inequality  and  less  in 
another,  the  inequalities  are  said  to  subsist  in  a  contrary  sense. 

a;  >  ffl  and  2/  >  ft  subsist  in  the  same  sense,  also  a;  <  3  and  y  <  4  ;  but  x  >  6 
and  y  <ia  subsist  in  a  contrary  sense. 

373 


374  INEQUALITIES 

400.  1.  If  2  is  added  to  each  member  of  the  inequality  8  >  5. 
how  will  the  two  inequalities  subsist  ?  How  will  they  subsist,  if 
2  is  subtracted  from  each  member  ? 

2.    Investigate  other  inequalities. 

Principle  1.  —  If  the  same  number  or  equal  numbers  are  added 
to  or  subtracted  from  both  members  of  an  inequality,  the  reciting 
inequality  will  subsist  in  the  same  sense. 

Let  a  >  6. 

Then,  §  397,  a  --  b  =  p,  a.  positive  number. 

Adding  c  —  c  =  0,  Ax.  2,     a  +  c  —  (b  +  c)  =  p. 
Therefore,  a  +  c>b  +  c. 

401.  1.  What  is  the  effect  of  adding  2  to  each  member  of  the 
inequality  x  —  2>y?  What  is  the  effect  of  subtracting  2  from 
each  member  of  the  inequality  a  +  2  >  6  ? 

2.  If  a  term  is  transposed  from  one  member  of  an  inequality 
to  the  other,  what  must  be  done  to  its  sign  ? 

Principle  2.  —  A  term  may  be  transposed  from  one  member 
of  an  inequality  to  the  other,  provided  its  sign  is  changed. 

Let  a  —  b>c  —  d. 

Adding  b  to  each  side,  Prin.  1,  a>b  +  c  —  d. 

Adding  —  c  to  each  side,  Prin.  1,  a  —  c  >  6  —  (?. 

402.  Principle  3.  —  If  the  signs  of  all  the  terms  of  an  inequality 
are  changed,  the  resulting  inequality  will  subsist  in  a  contrary  sense. 

Let  a  —  6  >  c  —  d. 

Transposing  every  term,  Prin.  2, 

—  c  +  d>  —  a  +  b; 
that  is,  — a  +  b<  —  c  +  d. 

403.  1.  If  both  members  of  the  inequality  10  >  8  are  multi- 
plied by  2,  how  will  the  two  inequalities  subsist  ?  How  will  they 
subsist,  if  both  members  are  divided  by  2  ? 

2.  How  will  they  subsist  in  each  case,  if  the  multiplier  or 
divisor  is  —  2? 


INEQUALITIES  375 

Principle  4.  —  If  both  members  of  an  inequality  are  multiplied 
or  divided  by  the  same  number,  the  resulting  inequality  ivill  subsist 
in  the  same  sense  if  the  multiplier  or  divisor  is  positive,  but  in  the 
contrary  sense  if  the  multiplier  or  divisor  is  negative. 

Let  a~>b. 

Then,  §  397,  a  —  b  =  p,  a.  positive  number. 

Multiplying  by  m,  ma  —  mb  —  mp. 

If  fli  is  positive,  mp  is  positive, 

and  ma  >  mb. 

If  m  is  negative,  mp  is  negative, 

and  mb  >  ma,  or  ma  <  mb. 

Putting  —  for  m,  the  principle  holds  also  for  division. 
m 

404.  1.  If  the  corresponding  members  of  the  inequalities  6  >  5 
and  4  >  2  are  added,  how  will  the  resulting  inequality  subsist  ? 
How,  if  —  5  >  —  6  and  —  2  >  —  4  are  added  ? 

Principle  5.  —  If  the  corresponding  members  of  any  number  of 
inequalities  subsisting  in  the  same  sense  are  added  together,  the 
resulting  inequality  ivill  subsist  in  the  same  sense. 

Let  a  >  b,  c'>d,  e  >/,  etc. 

Then,  §  397,  a  —  b,  c  —  d,  e  —  f,  etc.,  are  positive  numbers. 

Hence,  their  sum,  a  +  c  +  e+  •••  —  (b  ■\-  d-\-  f+  •••),  is  a  positive  number; 
that  Is,  a  +  c  + e +•••>/>  +  (?+/+ ••.. 

405.  Principle  6.  —  If  each  member  of  an  inequality  is  sub- 
tracted from  the  corresponding  members  ofa7i  equation,  the  resulting 
inequality  will  subsist  in  a  contrary  sense. 

Let  a  >  6  and  let  c  be  any  number. 

Then,  §  397,  a  —  6  is  a  positive  number. 

Since  a  number  is  diminished  by  subtracting  a  positive  number  from  it, 

c  —  (a  —  ft)  <  c. 

Transposing,  c  —  a<c  —  b. 

That  is,  if  each  member  of  the  inequality  «  >  ft  is  subtracted  from  the  cor- 
responding member  of  the  equation  c  —  c,  the  result  is  an  inequality  subsisting 
in  a  contrary  sense. 


376  INEQUALITIES 

406.  It  is  evident  that  the  difference  of  two  inequalities  sub- 
sisting m  the  same  sense,  or  the  sum  of  two  inequalities  subsist- 
ing in  a  contrary  sense,  or  the  product,  or  the  quotient  of  two 
inequalities,  member  by  member,  may  have  its  first  member 
greater  than,  equal  to,  or  less  than  its  second. 

For  example,  take  the  inequality  12  >  6. 

Subtracting  8  >  2,  or  adding  —  8  <  —  2,  the  result  is  4  =  4. 

Subtracting  8  >  1,  or  adding  —  8  <  —  1,  the  result  is  4  <  5. 

Multiplying  and  dividing  by  3  >  2,  the  results  are  36  >  12  and  4  >  3. 

Multiplying  by  —  2  >  —  4,  the  result  is  —  24  =  —  24.  Dividing  by  4  >  2, 
the  result  is  3  =  3. 

Multiplying  and  dividing  by  —  2  >  —  3,  the  results  are  —  24  <  —  18  and 
-  6  <  -  2. 

Examples 

407.  1.    Find  one  limit  of  x  in  the  inequality  3  a;  —  10  >  11. 

Solution 

3a;-10>ll. 

Prin.  2,  3x>21. 

Prin,  4,  x>7. 

Therefore,  the  inferior  limit  of  x  is  7 ;  that  is,  x  may  have  any  value  greater 
than  7. 

2.  Find  the  limits  of  x  in  the  simultaneous  inequalities 
3  a;  -f-  5  <  38  and  4  a;  <  7  ic  -  18. 

Solution 

3a;-f5<38.  (1) 

4x<7x-18.  (2) 

Transposing  in  (1),  Prin.  2,  3x<33. 

.-.Prin.  4,  a;<ll. 

Transposing  in  (2),  Prin.  2,        —  3x <  —  18. 

Changing  signs,  Prin.  3,  3  x  >  18. 

.-.Prin.  4,  x>6. 

The  inferior  limit  of  x  is  6,  and  the  superior  limit  is  11 ;  that  is,  the  given 
inequalities  are  satisfied  simultaneously  by  any  value  of  x  between  6  and  11. 


INEQUALITIES  377 

3.  Find  the  limits  of  x  and  ?/in3x— ?/>  — 14  and  x-\-2y  =  0. 

Solution 
3a;-y>-14.  (1) 

X  +  2  «/  =  0.  (2) 

Multiplying  (1)  by  2,  Prin.  4, 

6  X  -  2  y  >  -  28.  (3) 

Adding  (2)  and  (3),  7  x  >  -  28. 

.-.  x>-4.  (4) 

Multiplying  (2)  by  3,  3  x  +  6  y  =  0.  (6) 

Subtracting  (5)  from  (1),  Prin.  1, 

-7  2/>-14. 
Dividing  by  —  7,  Prin.  4,  y  <  2. 

That  is,  X  is  greater  than  —  4,  and  y  is  less  than  2. 

Find  the  limits  of  x  in  the  following : 

4.  6a;-5>13.  {4.x-ll>\x, 

5.  oa;-l<6a;  +  4.  ^*     l20-2a;>10. 

6.  3a;-|a;<30.  r3-4a;<7, 

7.  4a;  +  l<6a;-ll.  *     l5a;  +  10<20. 

10.   a;  +  — -f  — >25  and  <30. 
3        6 

11.    Find  the  limits  of  a;  in  ar^  +  3  cc  >  28. 

Solution 

x2  +  3  X  >  28. 
Transposing,  Prin.  2,  x2  +  3  x  -  28  >  0. 

Factormg,  (x  -  4)  (x  +  7 )  >  0. 

That  is,  (x  —  4)(x  +  7)  is  positive. 

If  (x  —  4)  (x  +  7)  is  positive,  either  both  factors  are  positive  or  both  are 
negative.  Both  factors  are  positive,  when  x  >  4  ;  both  factors  are  negative, 
■vhen  X  <  -  7. 

Therefore,  x  can  have  all  values  except  4  and  —  7  and  intermediate  values. 


378  INEQUALITIES 

Find  the  limits  of  x  in  the  following : 

12.  a^  +  3a:>10.  16.    a^>9a;-18. 

13.  a;2  +  8ar>20.  17.    cc^  +  40a;  >  3(4a;- 25). 

14.  y?  -{-5x>  24.  18.    x^  ->rhx>ax  +  ah. 

15.  (a;  -  2)  (3  -  a;)  >  0.  19.    (a;  -  3)  (5  -  a;)  >  0. 

Find  the  limits  of  x  and  y  in  the  following,  and,  if  possible, 
one  positive  integral  value  for  each  unknown  number : 

2x-3y<2,  jy  =  3a;  +  4, 

■       2a;  +  52/  =  25.  '     \25<2y  +  Zx. 

3x  +  2?/  =  42,  fy  — a;>9, 

3x     ^>lb.  [20^15-^- 

x  +  ?/  =  10,  f  X  >  w  +  4, 

22.-1  25.     1      ^^^    ' 

4x<3t/.  la;-2^  =  8. 

If  a,  b,  and  c  are  positive  and  unequal, 

26.  Which  is  the  greater,  — — —  or  — i ? 

^  '  a  +  26        a  +  3& 

27.  Prove  that  a^  +  b->2  ah. 

Suggestion.  —  Whether  a  —  6  is  positive  or  negative,  (o  —  ft)*  is  positive. 

28.  Prove  that  a^  +  W  +  c?>ah  +  etc  -f  he. 

29.  Prove  that  a^  +  6^  >  a^6  +  a6l 

30.  Which  is  the  greater,  ^^d^  or  ^^!±_^  ? 

a  +  6        a-  +  6^ 

31.  Prove  that 1 >  1,  except  when  2  a  =  3  6. 

3&     4a 

32.  Prove  that  (a  —  2  &)  (4  &  —  a)  <  fe^,  except  when  a  =  3  6. 

33.  Prove  that  a^  +  ft''  4-  c^  >  3  a&c. 

34.  Prove  that  the  sum  of  any  positive  real  number,  except  1, 
and  its  reciprocal  is  always  greater  than  2. 


VARIABLES   AND   LIMITS 


408.  A  number  that  has  the  same  value  throughout  a  discussion 
is  called  a  Constant. 

Arithmetical  numbers  are  constants.  A  literal  number,  as  a  or  x,  is  con- 
stant in  a  discussion,  if  it  keeps  the  same  value  throughout  that  discussion. 

409.  A  number  that  under  the  conditions  imposed  upon  it  may 
have  a  series  of  different  values  is  called  a  Variable. 

Variables  are  usually  represented  by  x,  y,  z,  etc. 

The  numbers  .3,  .33,  .333,  .3333,  .  .  .  are  successive  values  of  a  variable 
approaching  the  constant  |. 

The  commensurable  numbers  1,  1.4,  1.41,  1.414,  1.4142,  .  .  .  are  succes- 
sive values  of  a  variable  approaching  the  constant  V2. 

410.  An  expression  whose  value  depends  upon  the  value  of  a 
variable  is  called  a  Function  of  that  variable. 

2  a;  +  1  is  a  function  of  x,  because,  if  successive  values  are  given  to  x, 
2  X  +  1  will  take  successive  values.     For  example,  if  x  =  0,  2  x  +  1  =  1 ;  if 

x  =  l,  2x-|-l  =  3;    ifx  =  2,    2x  +  l  =  5,  etc.    x^  —  2 x  and    are    also 

functions  of  x.  ~^ 

x^  +  2  X2  —  5  z2  is  a  function  of  x  and  z. 

Every  function  of  a  single  variable  is  a  variable. 

The  variable  to  which  different  values  may  be  given  at  pleasure 
or  according  to  some  law  is  called  the  Independent  Variable,  and 
the  function  of  the  independent  variable  is  called  the  Dependent 
Vanable. 

In  the  first  illustration  x  is  the  independent  variable  and  2  x  -|- 1,  the  func- 
tion of  X,  is  the  dependent  variable. 

411.  To  discuss  functions  of  a  variable  it  is  necessary  to  sup- 
pose that  the  variable  takes  its  successive  values  according  to 
some  definite  law  of  change. 

379 


380  VARIABLES  AND  LIMITS 

Where  a  variable  takes  a  series  of  values  that  approach  nearer 
and  nearer  a  given  constant,  so  that  by  taking  a  sufficient  number 
of  steps  the  difference  between  the  variable  and  the  constant  can 
be  made  numerically  less  than  any  conceivable  number  however 
small,  the  constant  is  called  the  Limit  of  the  variable,  and  the 
variable  is  said  to  approach  its  limit. 

The  variable  .3,  .33,  .333,  .3333,  .  .  .,  whose  increase  at  each  step  is  y^  of 
the  previous  increase,  approaches  J  as  a  limit.  For  .3  differs  from  ^  by  less 
than  yJj,  .33  by  less  than  y^j,  .333  by  less  than  ys\^,  etc.,  and  by  taking  a 
sufficient  number  of  steps  it  is  possible  to  obtain  a  value  of  the  variable  differ- 
ing from  ^  by  less  than  any  number  that  can  be  conceived  of,  however  small. 

412.  The  difference  between  a  variable  and  its  limit  is  a  vari- 
able whose  limit  is  zero. 

As  .3,  .33,  .333,  .  .  .  approaches  its  limit  ^,  the  variable  difference  |  —  .3, 
^  —  .33,  ^  —  .-363,  .  .  .,  or  ^,  5^5,  ^^Viy  •  •  •  approaches  the  limit  zero. 

A  variable  may  approach  a  constant  without  approaching  it  as 
a  limit. 

The  variable  6.6,  6.66,  6.666,  .  .  .,  in  approaching  6§  as  a  limit,  approaches 
7  also,  but  not  as  a  limit. 

A  variable  in  approaching  its  limit  may  be  always  less  than  its 
limit,  or  always  greater,  or  sometimes  greater  and  sometimes  less. 

The  variable  .3,  .33,  .333,  ...  is  always  less  than  its  limit  |. 
The  variable,  .7,  .67,  .667,  ...  is  always  greater  than  its  limit  1  —  |. 
/       The  sum  of  the  first  n  terms  of  the  geometrical  series  1,  —  J,  J,  —  5,  ^^, 
—  3^2,  ...  is  a  variable  whose  successive  values  1,  ^,  |,  f,  |^,  f^,  .  .  .  are 
alternately  greater  and  less  than  the  limit  |. 

A  variable  may  increase  or  decrease  according  to  its  law  of 
change  and  become  numerically  greater  or  less  than  any  assign- 
able number. 

The  variable  1,  —2,  4,  —  8,  16,  .  .  .  may  become  numerically  greater  than 
any  number  that  can  be  assigned.  The  variable  1,  ^,  \,  \,  ^^,  .  .  •  may 
become  numerically  less  than  any  number  that  can  be  assigned. 

413.  A  variable  that  may  become  numerically  greater  than  any 
assignable  number  is  said  to  be  Infinite. 

The  symbol  of  an  infinite  number  is  00. 

414.  A  variable  that  may  become  numerically  less  than  any 
assignable  number  is  said  to  be  Infinitesimal. 

An  infinitesimal  is  a  variable  whose  limit  is  zero. 


VARIABLES  AND   LIMITS  381 

The  character  0  is  used  as  a  symbol  for  an  infinitesimal  number 
as  well  as  for  absolute  zero,  the  result  obtained  by  subtracting  a 
number  from  itself. 

415.  A  number  that  cannot  become  either  infinite  or  infinitesi- 
mal is  said  to  be  Finite.    / 

416.  Interpretation  of  -• 

If  the  numerator  of  the  fraction  -  is  constant  while  the  de- 

X 

nominator  decreases  regularly  until  it  becomes  numerically  less 
than  any  assignable  number,  the  quotient  will  increase  regularly 
and  become  numerically  greater  than  any  assignable  number. 

.-.  ^  =  00.     That  is, 

If  a  finite  number  is  divided  by  an  infinitesimal  number,  the 
quotient  tvill  be  an  infinite  number. 

417.  Interpretation  of  — 

00 

If  the  numerator  of  the  fraction  -  is  constant  while  the  de- 

X 

nominator  increases  regularly  until  it  becomes  numerically  greater 
than  any  assignable  number,  the  quotient  will  decrease  regularly 
and  become  numerically  less  than  any  assignable  number. 

.-.  -  =  0.     That  is, 

00 

If  a  finite  munber  is  divided  by  an  infinite  number,  the  quotient 
will  be  an  infinitesimal  number'. 

418.  The  symbol  =  is  read  'approaches  as  a  limit.' 
The  abbreviation  lim.  is  used  for  limit. 

X  =  a  is  read  'x  approaches  a  as  its  limit.'  The  expression  lira,  x  =  a  ia 
equivalent  to  x  =  a,  and  is  read  '  the  limit  of  a;  is  a.' 

Though  00  represents  a  variable  that  may  transcend  all  finite 
values,  it  is  convenient  to  use  the  expression  a;  =  oo  to  indicate 
that  X  increases  numerically  without  limit. 

Thus,  as  «  =  00,   -  :i  0. 

•  ■        X 

419.  Principle  1, — A  variable  cannot  approach  two  unequal, 
limits  at  the  same  time. 


382  VARIABLES  AND  LIMITS 

For  in  approaching  the  more  remote  as  a  limit  the  variable  would  reach  a 
value  intermediate  between  the  two,  and  thereafter  in  approaching  one  as 
a  limit  it  would  recede  from  the  other,  which,  therefore,  could  not  be  a  limit. 

420.  Principle  2.  —  If  two  variables  are  always  equal  and  each 
approaches  a  limit,  their  limits  are  equal. 

For  in  their  values  the  two  variables  are  but  one. 

Hence,  Prin.  1,  the  limit  of  their  common  values  is  their  common  limit. 

421.  1.  Since  the  limit  of  .333  •••is  i,  what  is  the  limit  of 
2  +  .333  •••,  or  2.333  •••  ?  of  4  +  .333,  or  4.333  ...  ?  of  5.333  ...  ? 
of  2  -  .333  ...,  or  1.666  ...  ?  of  .333 .13,  or  .203  ...  ? 

2.  What  is  the  limit  of  the  algebraic  sum  of  a  constant  and  a 
variable  ? 

Principle  3.  —  TTie  limit  of  the  algebraic  sum  of  a  constant  and 
a  variable  is  the  algebraic  sum  of  the  constant  and  the  limit  of  the 
variable. 

422.  1.  Since  the  limit  of  .333  ...  is  i,  what  is  the  limit  of 
.666...?  of  .111  ...  ? 

2.  Since  the  limit  of  1  +  i  +  i  -f  ...  is  2,  what  is  the  limit  of 

•^  4-  3  _i_  3     I     , . ,  ?    of  J     -4-    L  -I-     1     4-  . . .  '.> 

3.  How  may  the  limit  of  the  product  of  a  variable  and  a  con- 
stant be  obtained  from  the  limit  of  the  variable  ? 

Principle  4.  —  The  limit  of  the  product  of  a  variable  and  a  finite 
constant  is  equal  to  the  ^product  of  the  constant  and  the  limit  of  the 
variable. 

The  above  principle  may  be  established  as  follows : 

Case  1.  —  When  the  limit  of  x  is  0. 

Let  k  be  any  finite  constant. 

It  is  to  be  proved  that  kx  ~  0. 

However  small  any  number,  as  q,  may  be,  since  a:  =  0,  x  may  be  made 
numerically  less  than  q  -^  k. 

Hence,  kx  may  be  made  numerically  less  than  q  ;  that  is,  kx  may  be  made 
numerically  less  than  any  number  Jiowever  small. 

Therefore,  §411,  kx  =  0. 

Case  2.  —  Whe7i  the  limit  of  x  is  not  0. 

Let  k  be  any  finite  constant,  a  the  limit  of  x,  and  y  the  variable  that  must 
be  added  to  x  to  produce  a. 


[VARIABLES  AND  LIMITS  383 

It  is  to  be  proved  that  kx  =  ka. 

Since  x  +  2/  =  a,  x  =  a  —  y; 

.-.  kx  =  ka  —  ky. 

Prin.  2,  lim.  {kx)  =  lim.  {ka  —  ky) 

Prin.  3,  =  ka  —  lim.  {ky). 

But  since,  §  412,  y  =  0,  by  Case  1,  lim.  ky  =  0. 

Hence,  lim.  {kx)  —  ka  —  0  =  ka ; 

that  is,  kx  =  ka. 

Note.  —  The  principle  holds  for  the  limit  of  a  variable  divided  by  a  con- 
stant, since  dividing  by  k  is  equivalent  to  multiplying  by  -• 

423.  Principle  5.  —  The  limit  of  the  variable  sum  of  any  finite 
number  of  variables  is  equal  to  the  sum  of  their  limits. 

The  above  principle  may  be  established  as  follows : 

Let  a;  =  a,  2/  =  6,  2  =  c,  etc.,  to  any  finite  number  of  variables,  as  n. 

It  is  to  be  proved  that  lim.  {x  -{■  y  +  z  -\-  •••)  =:  a  +  6  +  c  +  ■••. 

Let  t?i,  ij2i  ^3,  •••  be  the  variable  differences  between  x,  y,  z  •••  and  their 
respective  limits. 

Then,        a;  +  y  +  2  +  •••  =  (a  +  6  +  c  +  •••)  -  {vx  +  V2  +  »3  +  •••)> 
and      lim.  (x  +  y  +  z  +  ■.•)  =  lim.  [(«  +  6  +  c  +  •••)  —  {vx  +  V2  +  «3  ^ — )] 

Prin.  3,  =  a  +  6  +  c  +  •••  —  lim.  (»i  +  W2  +  »3  +  •••)• 

Since,  §412,  »i  =  0,  ^2  ^  0,  v^  =  0,  etc.,  however  small  any  number,  as 
g,  may  be,  each  of  the  n  variables,  vi,  V2,  vg,  etc.,  may  be  made  less  than 
q  H-  n,  and  hence  their  sum  may  be  made  less  than  q. 

Therefore,  §411,         lim.  {vi  +  V2  +  vs  +  ••■)  =  0. 

Hence,  lim.  {x  +  y  +  z  +  ••■)  =  a  +  b  +  c  +  •••. 

424.  Principle  6.  —  The  limit  of  the  variable  product  of  two 
or  more  variables  is  equal  to  the  product  of  their  limits. 

The  above  principle  may  be  established  as  follows : 

Given,  x  =  a  and  y  =  b. 

It  is  to  be  proved  that  lim.  {xy)  =  ah. 
Let  t7i  =  a  —  «  and  V2=  b  —  y. 

Then,  xy  =  {a  —  vi)  {b  —  v^), 

and  lim.  {xy)  =  lim.  [ab  —  bvi  —  av^  +  v^v^l 

Prin.  5,  Prin. 4,  =ab  —  b  lim.  vt  —  a  lim.  v^  +  lim.  {viVa) 

§  412,  =ab  +  lim.  {viv^). 


384  VARIABLES  AND  LIMITS 

Since  when  vi  and  v^  are  near  their  common  limit  0,  their  product  is  much 
less  than  either  vi  or  v^, 

lim.  (viVz)  =  0. 
Hence,  lim.  {xy)  =  ab  +  0  =  ab. 

Similarly,  the  principle  may  be  established  for  any  number  of  variables. 

425.  Principle  7.  —  The  limit  of  the  variable  quotient  of  two 
variables  is  equal  to  tJte  quotient  of  their  limits,  provided  the  limit 
of  the  divisor  is  not  0. 

The  above  principle  may  be  established  as  follows: 

V 
Let  X  =  -  :  and  let  Mm.  z  be  not  0. 

2 

It  is  to  be  proved  that         lim.  x  =         ^. 

lira,  z 

Since  x  =  -,  v  =  xz. 

z 
Prin.  6,  lim.  y  =  lim.  a;  •  lim.  z. 

Therefore,  lim.  x  =  ^-i!^^. 

lim.  z 

The  principle  has  no  meaning  when  ^  =  0,  since  lim.  y  cannot  be  divided 
byO. 

426.  When  by  causing  a  variable  x  to  approach  sufficiently 
near  to  a  it  is  possible  to  make  the  value  of  a  given  function  of  x 
approach  as  near  as  we  please  to  a  finite  constant  I,  I  is  called  the 
limit  of  the  function  when  x  =  a. 

Suppose  that  .1,  .11,  .111,  .1111,  ...  are  successive  values  of  x  approaching 
I  as  a  limit.  Then,  the  corresponding  values  of  1  —  2a;,  a  function  of  x,  are 
.8,  .78,  .778,  .7778,  values  of  a  variable  approaching  ^  as  a  limit.  As  a;  =  J, 
the  function  of  x,  1  —  2x,  =  | ;  for  by  causing  x  to  approach  sufficiently 
near  to  \  it  is  possible  to  make  1  —  2  x  approach  as  near  as  we  please  to  |. 

The  expression 

lim.  [function  of  x\^a 

is  read  '  limit  of  function  of  x  as  a;  approaches  a  as  a  limit.' 

Thus,  lim.  (1  —  2x)x=\  —  I  indicates  that  as  x  approaches  its  limit  |, 
1  —  2  X  approaches  the  limit  |. 

427.  In  finding  the  limiting  values  of  the  functions  given  in 
the  following  examples,  the  student  is  expected  to  apply  the 
principles  that  have  been  established  above. 

Finding  the  limiting  value  of  a  function  of  a;  as  x  =  a  is  called 
evaluating  the  function  for  x  =  a. 


VARIABLES  AND  LIMITS 


385 


Examples 
If  x  =  a,  y  =  2,  and  z  =  0,  find  the  limit  of 

1.  x  +  y  +  z.  A.    x-\y+ax-y^. 

2.  axy-x^.  5.    {x  +  y)x -{x  -  y)z. 


o   ^    y. 

2      a 


^     a+x+z  ^ x+y 
x  —  y  a 


Find  the  value  of 


7.   Lim.r^-^^  +  »' 
X  —  2 


x=3 


8.  Lim. 

9.  Lim. 
10.    Lim. 


fa^  +  4  a;  +  1" 
L  X*  +  a^  4- 1  _ 

V  -  3  a;  +  4' 

aj  +  e 

"1  +  a;"" 
1  4- a; 


j;=-l 


£=2 


z=\ 


428.    Indeterminate  forms   -,    — ,    0  x  qo,    qo  x  0,    oo  —  oo. 

0       00 

For  all  values  of  a  and  x, 
a 

X  _a    X 
l~l  'i 

X 


=  a. 


If  a;  =  00,  (1)  becomes    -  =  0  x  oo  =  a. 


(1) 


If  a;  =  0,  (1)  becomes    -  =  oo  x  0  =  a. 

00 

Since  a  denotes   any  number  whatever,    -,    — ,    Ox  oo,   and 

0        00 

00  X  0  are  symbols  for  indeterminate  numbers. 

If  k  is  any  constant,  oo,  +  A;  =  0C2 ;    ••.  002  —  ooj  =  A;. 
Hence,  00  —  00  is  a  symbol  for  an  indeterminate  number. 

ADV.    ALG. — 25 


886  VARIABLES  AND  LIMITS 

429.  Since  every  function  of  a  single  variable  is  a  variable,  it 
is  evident  that  the  preceding  principles  apply  to  functions  of  a 
variable.  Thus,  to  apply  to  functions  of  a  variable,  Prin.  5,  6,  and 
7  may  be  stated  as  follows : 

Tlie  lirfiit  of  the  sum  of  a  finite  number  of  functions  of  x  is  equal 
to  the  sum  of  their  limits. 

The  limit  of  the  product  of  a  finite  number  of  functions  of  x  is 
eqxial  to  the  product  of  their  limits. 

Tlie  limit  of  the  quotient  of  two  fxmctions  of  x  is  equal  to  the 
quotient  of  their  limits,  provided  the  limit  of  the  divisor  is  not  zero. 

These  principles  fail  to  give  a  limit  whenever  the  result 
obtained  involves  one  of  the  indeterminate  forms, 

00  —  00,  0  X  00,  QO  X  0,  -,  ^• 
'  '  '  Q'  CO 

430.  The  preceding  principles  of  limits  lead  to  the  conclusion 
that  the  limit  of  a  function  is  found  by  substituting  the  limits  of 
the  variables  for  the  variables,  except  when  such  a  substitution 
gives  an  indeterminate  form  (§  429). 

Thus,  if  lim.  x  =  5  and  lim.  ?/  =  2,  the  limit  of  4  a;  —  3  y  is  found  by  substi- 
tuting 5  for  X  and  2  for  y  in  the  function  ix  —  3y.    But  if  substitution  is 

employed  directly  to  evaluate  the  functions 7^,  -  x  y,  and 

X  —  I     x{x  —  \)y 

(a;2  —  1)  -^  (x  —  1)  when  x  =  1  and  y  =  0,  these  functions  take  the  forms 

00  —  00  ,  00  X  0,  and  0-4-0,  respectively. 

When  the  method  of  evaluation  by  substitution  in  the  given 
function  fails,  the  evaluation  of  the  function  is  performed  by  the 
aid  of  Prin.  2. 

Thus,  to  evaluate  (a-2  —  1)  ^(x  —  1)  when  x  =  1,  find  another  function  of 
X,  as  X  +  1,  equal  to  the  given  function  (x^  —  1)-t-(x—  1)  for  all  values 
assumed  by  x  while  approaching  the  limit  1. 

If  X  takes  the  successive  values 

2,  I,  I,  I,  Ih  11.  •••.  approaching  1, 
then  both  functions  (x^  —  1)  -r-  (x  —  1)  and  x  +  1,  take  the  successive  values 
3,  i,  I,  V'  il.  fl»  •••>  approaching  2. 
Since  the  two  functions  are  equal  for  all  values  of  x  as  x  approaches  its 
limit  1,  by  Prin.  2  they  have  the  same  limit.     This  limit  is  lim.  (x  +  l)xii, 
which  by  substituting  lim.  x  for  x  is  found  to  be  1  +  1,  or  2. 


VARIABLES  AND   LIMITS 


387 


Examples 


Find  the  value  of 


Lim. 


2.    Lim. 


fx"  -  81 

_a^-4 

x=2 

"1  -  ar'" 

J--X 

x  =  l 

5.    Lim. 

6.    Li 

m. 

3.    Lim. , 


4.    Lim.r?ii^" 


xp^— 6 


a^  —  5  a;  +  6" 
ar'-4 


-/:=2 


3a^  +  5ar'-2a;-l' 

a^  —  8  x^  +  7 


x=l 


7.    Find  the  limiting  value  of  — — — ~ ^t_  when  x  =  0 

and  also  when  a;  =  oo. 

Solution.  —  As  x  approaches  the  limit  0,  the  first  three  terms  of  the 
numerator  and  also  of  the  denominator  become  infinitely  small  as  compared 
with  the  fourth,  and,  consequently,  may  be  neglected.  Hence,  when  x  =  0, 
the  fraction  approaches  the  limiting  value  |. 

As  X  =  00,  that  is,  as  x  becomes  indefinitely  greater,  the  last  three  terms 
of  the  numerator  and  also  of  the  denominator  become  infinitely  small  as 
compared  with  the  first,  and,  consequently,  may  be  neglected.     Hence,  when 

X  =  cc,  the  fraction  approaches  the  limiting  value  — ^,  or  — 


Find  the  limiting  values  of  the  following  when  x  =  0  and  when 
X  =  00 : 


8. 


10. 


11. 


1+:^ 

2  +  X*  +  x« 

1  -  X-' 

-  X*  -  2  x« 

5x3- 

aj2  -1-  4  X  +  2 

2  x^  4-  3  .x^  -  X  -f- 1 

4:X* 

-3a^  +  x  +  l 

2x*- 

X^  —  X^  -{-  X  -{-  1 

2x*- 

3ar^  +  2x^-2 

2x2  +  x  +  l 


12. 


13. 


14. 


15. 


5x  +  10 

x2  +  2x  +  2 

3x-4 

x2  -  X  -  8 

2x2-4x  +  l 

2x-l 

4ar^  +  5a^  +  2x 

2x3  +  x-f  1 


INCOMMENSURABLE    NUMBERS 


431.  Though  an  incommensurable  number  (§  227)  cannot  be 
expressed  by  any  integer  or  by  any  fraction  with  integral  terms, 
a  commensurable  number  can  always  be  found  to  differ  from  any 
incommensurable  number  by  less  than  any  number  that  may  be 
assigned,  however  small. 

For  example,  though  V2  cannot  be  expressed  by  a  decimal  that 
terminates,  commensurable  numbers  can  be  found  to  approximate 
to  the  true  value  of  V2.  For  by  the  process  of  evolution  a  com- 
mensurable number  can  be  found  to  differ  from  V2  by  less, 
successively,  than  .1,  .01,  .001,  .0001,  •••,  the  difference  finally 
becoming  smaller  than  any  number,  however  small,  that  can  be 
assigned. 

To  generalize :  let  p  and  q  be  variables  each  of  which  may  take 
any  integral  value  whatever.     Then,  any  commensurable  number 

P 

may  be  exactly  represented  by  -,  and  since  q  may  be  made  as 

large  as  we  please  and  p  may  be  given  any  integral  value  what- 

P 
ever,  any  incommensurable  number  may  be  represented  by  -  to 

any  required  degree  of  approximation. 

Thus,  any  incommensurable  number  may  be  included  between 

p  p  -\-  \.  1 

-  and ,  each  of  which  differs  from  it  by  less  than  -;  and 

q  q     '  .     .     5        . 

since  q  may  be  made  as  large  as  we  please,  the  limit  of  this 

difference  is  zero. 

Hence,  an  incommensurable  number  may  he  regarded  as  the  limit 
of  a  variable  commensurable  mimher. 

Thus,  \/2  is  the  limit  of  the  series  1  +  ^  +  t^  +  rsW  +  "••'  ^^^^  ^"'"  °^  " 
terms  of  which  is  1,  1.4,  1.41,  1.414,  •••,  as  n  takes  the  successive  values  1,  2, 
3,  4,  .... 

888 


INCOMMENSURABLE   NUMBERS  389 

432.  By  §  429,  the  principle  of  §  420  may  be  stated  as  follows : 

If  two  functions  of  the  same  variable  or  variables  are  equal  for  all 
values  of  the  variables,  the  limits  of  the  two  functions,  as  the  variables 
approach  their  resjyective  limits,  are  equal. 

Thus,  X  +  y  and  y  -\-  x  are  equal  for  all  values  of  x  and  y,  and  if  x  and  y 
approach  limits,  lim.  (  x+  ?/)  =  lim.  (y  +  x). 

433.  Commutative,  Associative,  and  Distributive  Laws. 

Let  a,  b,  and  c  be  incommensurable  constants,  the  limits,  §  431, 
of  the  commensurable  variables,  x,  y,  and  z,  respectively. 

1.  Covimutative  Law  for  Addition  and  Subtraction. 

§  55,  x  +  y  :=  y  +x. 

Hence,  §  432,  lim.  (x  -\- y)  =  lim.  (y  +  x) ; 

that  is,  §  423,  a+b  =  b  +  a. 

Similarly,  a  —  6  =  —  &  +  a. 

2.  Commutative  Law  for  Multiplication  and  Division. 
§  82,  xy  =  yx. 

Hence,  §  432,  lim.  (xy)  =  lim.  (yx) ; 

that  is,  §  424,  ab  =  ba. 

Similarly,  §  106,  a^  -  =  -.  a. 

3.  Distributive  Law  for  Multiplication  and  Divisipn. 
§§  85,  87,  z  being  either  a  monomial  or  a  polynomial, 

(x-\-y)z  =  xz+  yz. 
Hence,  §  432,      lim.  \fx  +  y)z']  =  lim.  (xz  +  yz) ; 
that  is,  §§  424,  423,  (a  +  &)c  =  ac  +  be. 

Similarly,  §  106,  (a  +  6)i  =  a  •  i  +  6  •  - ; 

c  c  c 

that  is,  (a-{-b)-i-c  =  a^c-\-b-^c. 

4.  Associative  Laiv. 

The  Associative  Law  follows  from  the  Commutative  Law,  sub- 
ject to  the  rules  for  signs  given  in  §  68,  2  and  §  104,  2. 


390  INCOMMENSURABLE  NUMBERS 

434.    Meaning  of  a"  when  n  is  incommensurable. 

Since  an  incommensurable  number  is  neither  even  nor  odd,  if  a 
is  negative  or  imaginary,  the  rule  for  the  sign  of  a  power  given  in 
§  218,  Prin,  1,  cannot  be  used  to  find  the  sign  of  (a)"  when  n  is 
incommensurable.  In  the  following  discussion,  then,  a  will 
denote  a  positive  real  number. 

Since  V3  lies  between  corresponding  terms  of  the  two  series 
17    173    1732  p 


^'  10'  100'  1000' 


— ) 


18    174    1733  p  +  1 

and  2,  ^^,  ^^^,  ^^^,  -.,  -^-,  •.., 

it  may  be  inferred  that  a/^  lies  between  corresponding  terms  of 

a*,  a^',  a^^,  a^^,  •••,  o«,  •••, 

p+i 

and  a^,  a^^,  a^^^,  a^^^^,  ••-,  a  «  ,  •••, 

1 

provided  a'  is  restricted  to  indicate  the  principal  5th  root. 

Passing  to  the  general  case,  in  a"  let  n  be  the  incommensurable 

limit  of  the  variable  exponent  x. 

p  p  _|_  1 

Then  two  values  of  x,  as  —  and  ,  can  be  found  to  include 

'9  Q    ' 

n  between  them ;  and  by  taking  q  sufficiently  great  the  difference 

between  the  values  of  x,  and,  consequently,  the  difference  between 

n  and  either  value  of  x,  may  be  made  as  small  as  we  please. 

It  will  now  be  proved  that  the  difference  between  the  j^oivers 
p+i  p 

a  '  and  a',  and,  consequently,  the  difference  between  a"  and  either 
power,  can  be  made  as  small  as  we  please  by  taking  the  difference 
between  the  corresponding  exponents  sufficiently  small;  that  is,  that 

p+\       p 
lim.(a  «  —  a^\q^oo  —  0. 
1.    When  a  >  1.    . 

1                    1                                          I 
Since  a  >  1,  a»>  1  and  a»—  1  is  positive.     Let  a»—  1  =  z. 
1 
.-.  a»=F=  1+2!,  and  a=  (1  +  zy  =  1  +  qz  -\ , 

every  term  of  which  is  positive,  z  being  positive. 

Therefore,  a>l  +  qz,  whence  z  < , 

which  approaches  0  as  q-  =  00.  ^ 


INCOMMENSURABLE  NUMBERS  391 

1 
Hence,  when  a  >  1,  lim.  (a«  —  l)^=oo  =  0.  (1) 

Since  a^  <,a  when  q>p,  a«  remains  finite  as  q  =  cc.     Hence, 

p     1  p+i       p 

§  424  and  (1),    lim.  [a«(a«-  l)^q=oo  =  0,  or  lim.  (a  «  —  a«)g=oo  =  0. 

p+i       p 

2.  When  a  =  1,  a  «  -  a«=  0. 

3.  When  a  is  positive  and  <  1, 

In  this  case  ->  1  and  f  -  )  —  1,  or  a  «—  1,  is  positive. 
a  \aj 

Hence,  by  (1),       lim.  (a"'-  l)^^*  =  0.  (2) 

p+} 
Since  a  is  positive  and  <  1,  a  «   cannot  become  greater  than  1 
however  great  q  becomes.     Therefore,  §  424  and  (2), 
p+i     _i  p+i       p 

lim.  [—  a  «  (a  *—  l)]^=oc  =  0,  or  lim.  (a  «  —  a«)5ix  =  0. 

It  has  been  proved  that  lim.  (a  «  —  a«)gi  »  =  0  for  all  positive  real 

values  of  the  base  a.     This  formula  is  true  when  -  is  substituted 

1  .« 

for  a,  since  -  is  positive  and  real  when  a  is  positive  and  real, 
a 

p+i       _p 
Therefore,  lim.  (a    «  —  a  ^)q^!»  =  0, 

in  which  the  values  of  x  are  negative.     Hence, 

When  a  is  any  positive   real  base,  the   difference   between   two 

p+i  p  p+i 

values  of  a'  that  include  a"  between  them,  as  a  ^    and  a«,  or  a    « 

_p 
and  a  ',  can  be  made  as  small  as  we  please  by  taking  the  difference 

between    the    corresponding    commensurable    exponents    sufficiently 

small. 

Therefore,  when  n  is  incommensurable  and  a  is  positive  and 

real,  a"  is  defined  as  the  limit  of  a^  when  x  =  n, 

435.  When  the  change  in  a  function  of  a  variable  correspond- 
ing to  an  infinitesimal  change  in  the  variable  is  an  infinitesimal, 
the  function  is  called  a  Continuous  Function  of  the  variable,  or  the 
function  is  said  to  vary  continuously  with  the  variable. 

a*  is  a  continuous  function  of  x  when  a  is  positive  and  real. 


392 


INCOMMENSURABLE  NUMBERS 


436.   Proofs  of  the  laws  of  exponents. 

Let  m  and  n  be  incommensurable  constant  exponents,  and  a 
any  positive  real  number.  Then,  m  and  n  may  be  defined  by  the 
relations  x  =  m  and  y  =  n,  and  by  §  434,  a"  and  a"  may  be  defined 
by  the  relations  a'  =  a™  and  a"  =  a",  x  and  y  being  variable  com- 
mensurable numbers. 


I.     §  240, 

a'  ■a''  =  a'+v. 

(1) 

§432, 

lim.  {a'  •  a")  =  lim.  a*+». 

(2) 

§424, 

lim.  (ct" .  W)  =  (lim.  a^)  (lim.  a") 

by  definition. 

=  a"  •  a". 

(3) 

By  definition, 

lim.  a'+''  =  a^^". 

(4) 

By  (3)  and  (4), 

a"  •  a"  =  a^^". 

(5) 

II.     By  law  I, 

a"""  •  a"  =  a"'""''+"  =  a"*. 

a"-"  =  a"  ^  a". 

(6) 

III.     §  249, 

(a')''  =  a^''. 

(7) 

By  definition, 

(a')"  -  (a^)"  =  0  when  y  =  n, 

(8) 

and 

(a*)"  —  (a")"  =  0  when  x  =  m. 

(9) 

(8) +  (9),  §423, 

(ay  -  (a"*)"  =  0 ; 

that  is,  when  x  =  m  and  y  =  n,  lim.  (a')*  =  (a")". 

Again,  by  definition,       lim.  a'"  =  a"™  ^"^"^ 

ft'''"'  *  ^''"' ' 

=  a"". 
Since,  §  432,  by  (7),      lim.  (a'^y  =  lim.  a'*, 
by  (10)  and  (11),  (a™)"  =  a"". 

IV.  §  250,  Vaf  =  a'^, 

1  1 

or  .   (ay  =  a''. 

Hence,  by  the  preceding  proof, 

V.  §  251,  (aby  =  a'b'. 

§  432,  lim.  (aby  =  lim.  (a'J'). 

By  definition  and  §  424,     (aby  =  a^'b". 


(10) 


(11) 


INTERPRETATION   OF   RESULTS 


437.  When  the  roots  obtained  by  solving  an  equation  satisfy 
the  equation,  the  solution  has  been  properly  performed ;  but  the 
results  found  in  solving  a  ])7-oblem  may  sometimes  be  at  variance 
with  some  condition  of  the  problem.  Consequently,  the  inter 
pretation  of  results  becomes  important. 

POSITIVE  RESULTS 

438.  Since  the  numbers  sought  in  the  solution  of  a  problem 
are  arithmetical  rather  than  algebraic,  when  positive  results  are 
obtained,  it  is  not  likely  that  they  will  conflict  with  the  condi- 
tions of  the  problem.  Sometimes,  however,  even  a  positive  result 
violates  one  or  more  of  the  conditions  of  a  problem. 

In  such  cases  the  problem  is  impossible. 

439.  1.  A  club  consisting  of  25  members  raised  the  sum  of 
$  13  by  assessing  the  men  80  cents  each  and  the  women  40  cents 
each.     How  many  men  were  there,  and  how  many  women? 


Solution 

Let 

X  =  the  number  of  men. 

Then, 

2b  —  X  —  the  number  of  women  ; 

|a;  +  §(25-a;)=13; 

whence, 

x  =  7J, 

and 

26-x  =  17^. 

Though  the  numbers  found  will  satisfy  the  equation,  yet  since 
the  number  of  men  and  the  number  of  women  cannot  be  frac- 
tional, the  problem  is  impossible. 

393 


394  INTERPRETATION  OF  RESULTS 

2.  The  second  digit  of  a  number  expressed  by  two  digits  is 
twice  the  first,  and  4  times  the  first  digit  is  9  greater  than  the 
second.     What  is  the  number  ? 


Solution 

Let 

X  =  the  first  digit. 

Then, 

2  «  =  the  second  digit ; 

.•. 

4a;  =  2a;  +  9; 

whence, 

a;  =  4|,  the  first  digit, 

and 

2  X  =  9,  the  second  digit. 

While  these  numbers  satisfy  the  equation,  they  fail  to  satisfy 
the  implied  condition  that  the  digits  must  be  integers. 
Hence,  the  problem  is  impossible. 

NEGATIVE  RESULTS 

440.  A  few  examples  will  suggest  the  methods  to  be  employed 
in  the  interpretation  of  negative  results. 

1.   If  A  is  40  years  old  and  B  is  30,  in  how  many  years  will  A 

be  twice  as  old  as  B  ? 

Solution 

Let  X  =  the  number  of  years  after  which  A  will  be 

twice  as  old  as  B. 

Then,  40  +  a:  =  2  (30  +  a:)  ; 

whence,  x  =  —  20. 

Though  the  result  is  algebraically  correct,  inasmuch  as  —  20 
substituted  for  x  satisfies  the  equation,  nevertheless  it  is  arith- 
metically absurd.  Hence,  the  conditions  of  the  problem  are 
inconsistent  with  each  other.  Had  the  result  been  +  20,  the 
statement  that  A  would  be  twice  as  old  as  B  in  20  years  would 
have  been  arithmetically  reasonable.  However,  since  —x=+  20, 
the  equation  will  give  a  result  arithmetically  reasonable,  if  —  a;  is 
substituted  for  x ;  that  is,  if  x  is  taken  to  represent  the  number  of 
years  since  A  was  twice  as  old  as  B. 

The  conditions  of  the  problem  should,  therefore,  be  modified  as 
follows :  If  A  is  40  years  old  and  B  is  30,  how  many  years  ago 
was  A  twice  as  old  as  B  ? 


INTERPRETATION  OF  RESULTS  396 

2.    How  much  money  has  A,  if  \  of  his  money  is  5  dollars  more 
than  ^  of  it  ? 

Solution 

Let  X  =  the  number  of  dollars  A  has. 


Then, 


5-?  =  5. 
4     3 

Solving  X  =  -  60. 


While  the  result  —  60  satisfies  the  equation,  it  violates  the  sup- 
position, made  in  the  problem,  that  A  has  some  money. 
If  —  ic  is  substituted  for  x,  the  equation  becomes 

=  5 :  whence,  a;  =  60. 

3      4'  ' 

The  problem  when  modified  to  express  conditions  arithmetically 
reasonable  will  be :  How  much  money  has  A,  if  \  of  it  is  5  dollars 
less  than  ^  of  it ;  or,  if  —  60  dollars  is  interpreted  as  60  dollars  in 
debt :  How  much  money  does  A  owe,  if  \  of  what  he  owes  is  5 
dollars  more  than  ^  of  it  ? 

441.  From  the  above  discussions  we  may  infer: 

1.  A  negative  result  indicates  that  some  quantity  in  the  problem 
has  been  ajyplied  in  the  wrong  direction. 

2.  A  jxtssible  problem  analogous  to  the  given  j^roblem  may  be 
formed  by  changing  the  absurd,  conditions  to  their  opposites. 

Problems 

442.  Interpret  arithmetically  the  negative  results  obtained  by 
solving  the  following : 

1.  If  A  is  40  years  old  and  B  is  25,  in  how  many  years  will  B 
be  half  as  old  as  A  ? 

2.  Find  the  numbers  whose  sum  is  6  and  difference  10. 

3.  What  fraction  is  equal  to  f  if  1  is  added  to  its  numerator, 
or  to  I,  if  1  is  added  to  its  denominator  ? 

4.  A  boy  bought  some  apples  for  24  cents.  Had  he  received 
4  more  for  that  sum,  the  cost  of  each  would  have  been  1  cent  less 
How  many  did  he  buy  ? 


396  INTERPRETATION  OF  RESULTS 

5.  A  man  worked  7  days,  during  which  he  had  his  son  with 
him  3  days,  and  received  22  shillings.  He  afterwards  worked 
5  days,  during  which  he  had  his  son  with  him  1  day,  and  received 
18  shillings.     What  were  the  daily  wages  of  each  ? 

ZERO  RESULTS 

443.  When  the  result  obtained  by  solving  a  problem  is  zero,  it 
may  sometimes  indicate  that  the  problem  is  impossible,  and  some- 
times it  may  be  the  proper  answer  to  the  question. 

1.  A  dealer  had  two  kinds  of  tea  worth  75  and  60  cents  per 
pound,  respectively.  How  many  pounds  of  each  must  he  take  to 
make  a  mixture  of  45  pounds  worth  $27? 

Solution 
Let  X  =  the  number  of  pounds  of  the  better  kind. 

Then,  45  —  x  =  the  number  of  pounds  of  the  poorer  kind  ; 

.-.  I  X  +  1(45  -  x)  =  27  ; 
whence,  x  =  0. 

This  result  means  that  no  such  mixture  can  be  made.  In  fact, 
45  pounds  of  the  poorer  tea  is  worth  $  27. 

2.  A  is  48  years  old,  and  B  is  16  years  old.  After  how  many 
years  will  A  be  3  times  as  old  as  B  ? 

Solution 
Let  X  =  the  required  number  of  years. 

Then,  48  +  x  =  3(16  +  x). 

Solving,  X  =  0. 

This  result  indicates  that  A  is  now  3  times  as  old  as  B. 

3.  What  number  is  equal  to  the  square  of  itself  ? 

Solution 
Let  X  =  the  number. 

Then,  x  =  x^, 

x(;x  -  1)  =0; 

.-.  X  =  1  or  0. 
These  results  indicate  that  no  number  except  1  is  equal  to  the 
square  of  itself. 


INTERPRETATION  OF  RESULTS  397 

INDETERMINATE  RESULTS 

444.  1.  A  lady  being  asked  her  age  replied,  "If  from  3  times 
my  age  you  take  4  years  and  divide  the  difference  by  2,  you  will 
have  twice  my  age  less  half  of  my  age  4  years  hence."  What  was 
her  age  ? 

Solution 
Let  X  =  the  number  of  years. 

Then,  3^^  =  2x-^,  ^    (1) 

3a:-4  =  4x-x-4,  (2) 

(3-3)x  =  0;  (3) 

.-.   -  =  ?•  (4) 

Since  (2)  may  be  reduced  to  the  identity  3a;  —  4  =  3x  —  4,  it 
may  be  satisfied  by  any  value  of  a;  whatever.  This  relation, 
§  428,  is  expressed  by  (4).     Hence,  the  problem  is  indeterminate. 

Problems 

445.  1.  If  twice  a  certain  number  is  subtracted  from  the 
square  of  the  number,  the  result  will  be  1  less  than  the  square 
of  a  number  1  less.     What  is  the  number  ? 

2.  A  father  is  30  years  older  than  his  son,  and  the  sum  of  their 
ages  is  30  years  less  than  twice  their  father's  age.  What  is  the 
son's  age  ? 

3.  The  sum  of  the  first  and  third  of  three  consecutive  integers 
is  equal  to  twice  the  second.     What  are  the  integers  ? 

4.  A  bought  400  sheep  in  two  flocks,  paying  $1.50  per  head 
for  the  first  flock  and  $  2  per  head  for  the  second.  He  lost  30 
of  the  first  flock  and  56  of  the  second,  but  by  selling  the  rest  of 
the  first  flock  at  $  2  per  head  and  the  rest  of  the  second  at  $  2.50 
per  head,  he  neither  lost  nor  gained.  How  many  sheep  were 
there  in  each  flock  originally  ? 

5.  A  and  B  receive  the  same  monthly  salary.  A  is  employed 
10  months  in  the  year  and  his  annual  expenses  are  f  600.  B  is 
employed  8  months  in  the  year  and  his  annual  expenses  are  $  480. 
If  A  saves  as  much  money  in  4  years  as  B  saves  in  5  years,  what 
is  the  monthly  salary  of  each  ? 


398  INTERPRETATION  OF  RESULTS 


INFINITE  RESULTS 

446.    An  infinite  result  indicates  that  the  problem  is  impossible. 

1.  If  a  man's  yearly  income  is  a  dollars  and  his  yearly  ex- 
penses are  a  dollars,  in  how  many  years  will  he  have  saved  6 

dollars  ? 

Solution 

Let  X  =  the  required  number  of  years. 

Then,  x  =  — ^  =  -,  or  «. 

a  —  a     0 

That  is,  he  will  never  have  saved  h  dollars  in  this  way. 

2.  A  reservoir  is  fitted  with  three  pipes.  One  pipe  can  fill  the 
reservoir  in  15  hours,  the  second  can  fill  it  in  |  of  that  time,  and 
the  third  pipe  can  empty  it  in  6  hours.  If  the  reservoir  is  full 
and  the  three  pipes  are  opened,  in  what  time  will  it  be  emptied  ? 


Solution 

Let 

X  ■=  the  required  number  of  hours. 

Then, 

16      10 

1_1 
6     X 

Solving, 

60 
x  =  — ,  or  oo. 
0 

That  is,  the  reservoir  will  7iever  be  emptied  under  these  con- 
ditions. 


3.    What  numb 

3r  add 

Led  to  both  terms 

make  the  fraction 

equal 

tol? 

Solution 

Let 

X  =  the  number. 

Then, 

1  +  a;_j 

2  +  x 

Clearing, 

1  +  a;  =  2  +  x. 

Solving, 

x-lor--^; 
0         0    ' 

that  is, 

X  =+  CO  or  —  a 

Consequently,  there  is  no  such  number ;  but  the  larger  the 
number  in  numerical  value,  the  nearer  will  the  resulting  fraction 
approach  the  value  1. 


INTERPRETATION  OF  RESULTS  399 

THE  PROBLEM  OF  THE  COURIERS 

447.  Two  couriers,  A  and  B,  travel  on  the  same  road  in  the 
direction  from  X  to  F  at  the  rates  of  m  and  n  miles  an  hour, 
respectively.  At  a  certain  time,  say  12  o'clock,  A  is  at  P,  and  B 
is  at  Q,  a  miles  from  F.     Find  when  and  where  they  are  together. 

X. I F 

P  Q 

Solution 

Suppose  that  time  reckoned  from  12  o'clock  toward  a  later  time  is  positive, 
and  toward  an  earlier  time,  negative  ;  also,  that  distances  measured  from  P 
toward  the  right  are  positive,  and  toward  the  left,  negative. 

Let  X  represent  the  number  of  hours  from  12  o'clock,  and  y  the  number 
of  miles  from  P,  when  A  and  B  are  together.  Then,  they  will  be  together 
y  —  a  miles  from  Q. 

Since  A  travels  mx  miles  and  B  travels  nx  miles  before  they  are  together, 

y  =  mx,  (1) 

and  y  —  a  =  nx.  (2) 

Solving  (1)  and  (2), 

X  =  — - — ,  the  required  time.  (3) 


n 
ma 


,  the  required  distance.  (4) 


DiscnssiON 

1.  When  a>0  andm'^n. 

When  a  >  0  and  m  >  n,  the  numerator  and  denominator  in  (.3)  and  also  in 
(4)  are  positive ;  hence,  x  and  y  are  positive. 

That  is,  A  overtakes  B  some  time  after  12  o'clock,  somewhere  at  the  right 
of  P. 

2.  When  a  >  0  and  m<n. 

When  a  >  0  and  m  <  ?i,  both  x  and  y  are  negative. 

That  is,  at  12  o'clock  B  is  ahead  of  A  and  gaining  on  him,  and  they  were 
together  some  time  before  12  o'clock  and  somewhere  at  the  left  of  P. 

3.  When  a  >  0  and  m=  n. 

When  a  >  0  and  m  =  n,  x  and  y  are  positive  and  infinitely  great. 
That  is,  at  12  o'clock  B  is  ahead  of  A  and  traveling  at  the  same  rate ; 
consequently,  he  will  never  be  overtaken  by  A. 


400  INDETERMINATE  EQUATIONS 

4.  When  a  =  0  and  m  ^  n  07-  m  <^  n. 

When  a  =  0  and  m>  n  or  m <,  u,  x  —  0  and  y  =  0. 

Ifm>n,x  =  +0  and  y  =  -f-  0.  That  is,  at  12  o'clock  A  and  B  are  to- 
gether, and  A  is  passing  B. 

If  m  <  w,  X  =  -  0  and  y  =  -  0.  That  is,  at  12  o'clock  A  and  B  are  to- 
gether, and  B  is  passing  A. 

5.  When  a  =  0  and  m  =  n. 

When  a  =  0  and  m  =  n,  x  =  -  and  «  =  -  • 
0  ^0 

That  is,  A  and  B  are  together  at  12  o'clock,  and  since  they  travel  at  the 

same  rate  they  will  be  together  at  all  times. 


olOic 


INDETERMINATE    EQUATIONS 


448.  While  a  problem  that  presents  more  unknown  literal  num- 
bers than  independent  equations  involving  them  is  in  general 
indeterminate  (§  214),  yet  frequently  by  the  introduction  of  a 
condition  or  conditions  not  leading  to  equations,  the  number  of 
values  of  the  unknown  numbers  may  be  limited  and  these  values 
algebraically  determined.  A  common  condition  is  that  the  results 
shall  be  positive  integers. 

1.    Solve  the  equation  5x  -\-3y  =  35  in  positive  integers. 

Solution 

Since  x  and  y  are  positive  integers,  5  x  must  be  equal  to  5  or  a  multiple  of 
5,  and  3  y  must  be  equal  to  3  or  a  multiple  of  3.  Since  the  sura  of  these 
multiples  is  35,  if  the  multiples  of  5  are  subtracted  from  35,  one  or  more  of 
the  remainders  will  be  a  multiple  of  3,  if  the  problem  is  possible. 

The  only  multiples  of  5  that  subtracted  from  35  leave  multiples  of  3  are 
6  and  4  times  5. 

.•.  X  =  1  or  4  ;  whence,  y  =  10  or  5. 

Or,  since  x  must  be  a  positive  integer  and  by  transposition  5x  =  35  —  Sy, 
the  values  of  x  must  be  1,  2,  3,  4,  5,  or  6,  if  the  equation  is  possible.  Sub- 
stituting these  values  of  x  in  the  given  equation  and  rejecting  all  those  that 
give  negative  or  fractional  values  for  y,  the  positive  integral  values  are  found 
to  be  X  =  1  or  4,  and  y  =  10  or  6. 


INDETERMINATE   EQUATIONS  401 

2.    Solve  the  equation  5  a;  +  8  ?/  =  107  in  positive  integers. 


Solution 

bx  +  %y 

=  107. 

(1) 

Dividing  by  5,                             x  +  y  +^: 

5 

=  21  +  |. 
5 

(2) 

Collecting  integral  and  fractional  terms, 

x  +  y-2\: 

_2-3j/ 
5 

(3) 

Since  x  -\-  y 

-  21  is  integral,              ^~^^  = 
5 

=  an  integer. 

(4) 

If  2-3y_ 

2-5 
w,  an  integer,  then,  y  = -— 

— ,  which  is  in 

the 

fractional 

5  '  o    ,  ,  .  3 

form.     To  avoid  this,  the  coefl&cient  of  y  in  the  number  placed  equal  to  w 

2  —  3  u 

should  be  made  equal  to  unity.     Since  "  is  equal  to  an  integer,  any 

5 
multiple  of  it  is  equal  to  an  integer.     Since  5  is  contained  in  3  times  —  3  y, 
—  2  J/  times  with  a  remainder  of  y,  multiplying  (4)  by  3, 

6-  9m 

^  =  an  mteger 

=  l_22/  +  i±i^.  (6) 

5 

Then,  let  -JtJM.  =  x^^  an  integer.  (6) 

Solving  for  y,  y  =  bw  —  \.  (7) 

Substituting  in  (3),  z  =  23  -  8  w.  (8) 

Equations  (7)  and  (8)  are  called  the  general  solution  of  the  given  equation 
in  integers. 

To  make  y  and  x  positive  integers,  it  is  evident  from  (7)  that  we  must 
take  ro  >  0 ;  and  from  (8)  that  we  must  take  mj  <  3. 

Since  w  is  an  integer  greater  than  0  and  less  than  3,  w  =  1  or  2, 

When  to  =  1,  x  =  15,  «/  =  4  ; 

when  10  =  2,  a;  =    7,  y  =  9. 

3.  Determine  whether  the  equation  10  a; +  15  =  53  may  be 
satisfied  by  integral  values  of  x  and  y. 

Solution.  —  Dividing  by  5,  2  x  +  3  ?/  =  \5. 

If  X  and  y  are  integers,  the  first  member  is  integral. 

Since  the  first  member  is  equal  to  the  fraction  ^^/,  it  cannot  be  an  integer 
Hence,  x  and  y  cannot  be  integers  at  the  same  time ;  that  is,  the  equation 
is  not  satisfied  by  integral  values  of  x  and  y. 

ADV.  ALG.  — 26 


402  INDETERMINATE   EQUATIONS 

Solve  the  following  equations  in  positive  integers: 

4.    5x-f32/  =  49.  8.    12a;  +  5y  =  61. 

b.    3x  +  2y  =  b.  9.      Q>x  +  ly  =  72. 

6.  2a; +  7^  =  48.  10.     5x  +  9i/  =  75. 

7.  8a;  +  5.v  =  80.  11.      6a;  +  9?/  =  100. 

Find  the  least  integral  values  of  x  and  y  in  the  following: 

12.  2x  =  9  +  3?/.  14.    7a;- 2^  =  6. 

13.  5i/  =  2a;-|-7.  15.    5a;-3?/  =  l. 

f,  1        1  •         fa;  +  ?/4-2  =  6l 

16.  Solve  the  equations  i  ,,  \  va.  positive  integers. 

{2x-\-y  -z  =  l  ]       ^  ^ 

Solution 
x  +  y  +  z  =  6.  (1) 

2  X  +  y  -  z  =  7.  (2) 

Adding,  3  x  +  2  y  =  13.  (3) 

Solving  (3)  for  positive  integers,        x  =  3  and  y  =  2. 
Substituting  in  (1),  2  =  1. 

Solve  the  following  equations  in  positive  integers: 

[  a;  +  ?/  +  2;  =  8,  f  3  a;  +  2  ?/  =  17, 

17.  •!  ^        '  19.     I  ^  ' 
I  a;-?/ +  2  2  =  6.                               1^  +  22  =  14. 

18.  |2-  +  3,  +  .  =  15,  ^^^     iy  +  z  =  l, 
l3x  +  i/-2  =  8.  l3a;-2  =  7. 

21.  Separate  100  into  two  parts  one  of  which  is  a  multiple 
of  11,  and  the  other  a  multiple  of  6. 

22.  In  what  v/ays  may  a  weight  of  19  pounds  be  weighed  with 
5-pound  and  2-pound  weights  ? 

23.  A  man  has  $.300  that  he  wishes  to  expend  for  cows  and 
sheep.  If  cows  cost  $  45  apiece  and  sheep  $  6  apiece,  how  many 
can  he  buy  of  each  ? 

24.  If  9  apples  and  5  oranges  together  cost  52  cents,  what  is 
the  cost  of  one  of  each  ? 


INDETERMINATE   EQUATIONS  403 

25.  A  grocer  sold  two  packages  of  sugar  for  f  1.25.  One  pack- 
age contained  a  certain  number  of  pounds  of  7-cent  sugar,  the 
other  a  certain  number  of  pounds  of  5-cent  sugar.  How  many 
pounds  were  there  in  each  package  ? 

26.  A  man  sold  9  animals  —  sheep,  hogs,  and  cows  — for  $100. 
If  he  received  $  3  for  a  sheep,  $  6  for  a  hog,  and  $  35  for  a  cow, 
how  many  of  each  did  he  sell  ? 

27.  A  woman  expended  93  cents  for  14  yards  of  cloth,  some  at 

5,  some  at  7,  and  the  rest  at  10  cents  a  yard.     How  many  whole 
yards  of  each  did  she  buy  ? 

28.  Divide  74  into  three  parts  that  shall  give  integral  quotients 
when  divided  by  5,  6,  and  7,  respectively,  the  sum  of  which 
quotients  shall  be  12. 

29.  A  purse  contained  30  coins,  consisting  of  half-dollars, 
quarters,  and  dimes.  How  many  coins  of  each  kind  were  there, 
if  their  aggregate  value  was  f  6.50  ? 

30.  A  man  bought  100  animals  for  $99.  There  were  pigs, 
sheep,  and  ducks.  If  he  paid  $  6  for  a  pig,  $  4  for  a  sheep,  and 
50  cents  for  a  duck,  how  many  of  each  did  he  buy  ? 

31.  What  is  the  least  number  that  will  contain  25  with  a 
remainder  of  1,  and  33  with  a  remainder  of  2  ? 

32.  Find  the  least  number  that  divided  by  10  and  by  11  will 
leave  remainders  of  3  and  6,  respectively. 

33.  What  is  the  least  number  that  will  contain  2,  3,  4,  5,  and 

6,  each  with  a  remainder  of  1,  and  7  without  a  remainder  ? 

34.  A  man  selling  eggs  to  a  grocer  took  them  out  of  his  basket 
4  at  a  time  and  there  was  1  egg  over.  The  grocer  put  them  into 
a  box  5  at  a  time  and  there  were  3  over.  Both  lost  the  count ; 
but  knowing  that  there  were  between  6  and  7  dozen  eggs,  the 
grocer  paid  for  6^  dozen.     How  many  eggs  did  he  lose  ? 

35.  Four  boys  have  a  pile  of  marbles.  A  throws  away  1  and 
takes  \  of  the  remainder ;  B  throws  away  1  and  takes  \  of  the 
remainder;  C  throws  away  1  and  takes  \  of  the  remainder;  D 
throws  away  1,  and  each  boy  takes  \  of  the  remainder.  At  least 
how  many  marbles  must  have  been  in  the  pile,  and  how  many 
does  each  boy  now  have  ? 


MATHEMATICAL  INDUCTION 


449.  Induction  is  generally  understood  as  the  process  of  infer- 
ring a  general  principle  from  particular  instances.  But  mathe- 
matical induction  diifers  in  some  respects  from  ordinary  induction. 
Mathematical  Induction  is  the  process  of  proving  a  general  prin- 
ciple by  means  of  a  known  fact  together  with  a  conditional 
principle.  The  known  fact  is  that  the  principle  under  consider- 
ation is  true  for  as  many  of  the  iirst  consecutive  cases  as  we 
examine,  and  the  conditional  principle  is  that  if  it  is  true  for  the 
wth  case  it  holds  true  for  the  (w  +  l)th  case. 

This  method  of  proof  is  illustrated  by  the  following  examples : 

1.  In  the  series  of  odd  numbers  1,  3,  5,  7,  •••,  the  first  odd 
number  =  2(1)  —  1 ;  the  second  odd  number  =  2(2)  —  1 ;  the 
third  odd  number  =  2(3)  —  1.  From  these  particular  instances 
it  may  be  inferred  that  the  «th  odd  number  =  2n  —  1,  and  it  is 
now  stated  as  a  principle  to  be  investigated  that  in  the  series 
1,  3,  5,  7,  •••,  any  odd  number  is  1  less  than  twice  the  number  of 
the  term. 

Supposing  that  the  wth  odd  number  is  2n  —  1,  by  the  law  of 
the  given  series  the  (n  +  l)th  odd  number  is  (2w  —  1)  -f  2. 

But,  §§  56,  85,  (2n  -  1)  +  2  =  (2n  +  2)  -  1  =  2(71  +  1)  -  1 ; 
that  is,  the  principle  holds  for  the  {n  -\-  l)th  odd  number  on  con- 
dition that  it  holds  for  the  nth  odd  number. 

Therefore,  since  the  principle  is  true  for  the  third  odd  number 
it  holds  for  the  fourth ;  since  it  is  true  for  the  fourth  it  holds  for 
the  fifth  ;  and  so  on.     Hence,  the  principle  is  true  generally'. 

2.  By  trial  x  —  y,  x-  —  y-,  ar'  —  i/^,  x*  —  y*,  and  x^  —  i/^  are  each 
found  to  be  divisible  by  x  —y.  From  these  special  cases  it  may 
be  inferred,  and  stated  as  a  principle  to  be  proved  or  disproved, 

404 


MATHEMATICAL   INDUCTION  405 

that  the  difference  of  any  like  powers  of  two  numbers  is  divisible 
by  the  difference  of  the  numbers.  But  the  proved  fact  that  x  —  y, 
or  —  y"^,  ••• ,  ar*  —  ?/*  are  each  divisible  hj  x  —  y  is  no  warrant  for 
accepting  without  further  inquiry  the  statement  that  ic"  —  y^,  for 
example,  is  divisible  by  x  —  y.  It  is  first  necessary  to  know 
whether  a^  —  y^  can  be  expressed  in  terms  of  these  numbers 
known  to  be  exactly  divisible  by  x  —  y.  Since  x^  —  y^  =  x^  —  x^y 
+  x^y  —  if  =  afix  —  y)  -\-  y{x'  —  y^),  each  term  of  which  has  been 
proved  to  be  divisible  hy  x  —  y,  x^  —  y^  is  divisible  hy  x  —  y;  and 
this  illustrates  the  general  proof. 

For  since       a;"+'  —  ?/""^^  =  a;"+'  —  x^y  -\-  x"//  —  y"+^ 
=  x\x-y)-]-y{x"  -y"), 

which  is  divisible  by  x  —.y  on  condition  that  ic"  —  ?/"  is  divisible 
by  a;  —  y,  it  follows  that  a;"+^  —  ?/"+'  is  divisible  hy  x  —  y  on  condi- 
tion that  x"  —  I/"  is  divisible  hy  x  —  y. 

These  two  things  then  are  proved : 

First,  the  difference  of  the  same  jwwers  of  any  two  numbers,  up 
to  and  including  the  fifth  powers,  is  divisible  by  the  difference  of  the 
numbers. 

Second,  if  the  difference  of  the  nth  powers  of  any  two  numbers  is 
divisible  by  the  difference  of  the  numbers,  the  difference  of  the  next 
higher  jwivers  is  divisible  by  the  difference  of  the  numbers. 

Therefore,  since  by  actual  division  a^  —  y^  has  been  proved  to 
be  divisible  by  x  —  y,  x^  —  y^  is  divisible  hy  x  —  y;  since,  as  just 
proved,  x^  —  y*^  is  divisible  hy  x  —  y,  x^  —  y''  is  divisible  by  a;  —  y ; 
since,  as  just  proved,  x^  —  y^  is  divisible  by  x—y,  x?  —  y^  is 
divisible  hy  x  —  y;  etc. 

Hence,  for  all  positive  integral  values  of  n,  however  great, 
»"  —  y"  is  divisible  hy  x  —  y. 

3.  Let  it  be  required  to  derive  a  formula  for  squaring  any 
polynomial. 

By  actual  multiplication,  rearranging  terms, 
(a  +  6f  =  a2  +  62^_2a6;  (1) 

(a  +  &  -f  c)2  =  a^  +  62  +  c^  +  2  a&  +  2  ac  +  2  6c;  (2) 

(a  +  6  +  c  +  («)'  =  a^  +  6^  +  c^  +  d2 

+  2  a6  +  2  ac  +  2  ad  +  2  6c  +  2  6d  +  2  cc?.  (3) 


406  MATHEMATICAL   INDUCTION 

It  is  proved,  then,  that  the  square  of  a  polynomial  of  n  terms, 
provided  that  n  is  not  greater  than  4,  is  equal  to  the  sum  of  the 
squares  of  the  terms,  plus  twice  the  product  of  each  term  by 
each  term  that  follows  it.  It  may  be  inferred  that  this  principle 
is  of  general  application,  but  without  further  proof  it  does  not 
follow  that  it  holds  for  polynomials  of  more  than  four  terms. 
'  Suppose,  however,  that  the  principle  is  true  for  a  polynomial 
of  n  terms,  n  being  any  positive  integer.     If  this  is  so,  then, 

(Oi  +  0-2  +  03+  •■•  +  (/„)-  =  af  +  a|  +  a|  +  •••  +  < 

+  2  «i(a2  +  rtg  +  •••  +  a„)+  2  a^{a^  +  04  +  •••  +  a„) 
+  ...+2a„_ia„  (4) 

is'the  formula  for  the  square  of  a  polynomial  of  n  terms. 

Since  {a^  +  a,  +  013+ \-  a„+i)-  =  [a^+ia^+a^  + \-  a^^-^J, 

by  (1),    (ai  +  a,  +  03  •••  +  a„+i)' 

=  a\+  {a-i  +  03  H +  a„+i)-  +  2  a,(a2  +  0,3  +  •••  +  a„+i) ; 

if  (4)  is  true, 

=  of  +  [a|  +  a|  +  •••  +  a^^i  +  2  a^ia^  +  •••  +  a„+i) 
+  203(04+  •••  +o„+i)+  •••  +2a„a„+i] 

+  201(02  +  03+ h  a„+i); 

rearranging, 

=  af  +  o^  +  o?  +  •••  +  o^+i  +  2  ai(o2  +  O3  +  •••  a„+i) 
+  2  02  (03  +  •••  +  a„+i)  +  2  03(04  +  •••  +  o„^i) 
+  ••• +2a„o„+i.  (5) 

But  (5)  expresses  the  same  law  for  forming  the  square  of  a 
polynomial  of  »  +  l  terms  that  (4)  expresses  for  forming  the  square 
of  a  polynomial  of  n  terms.  Hence,  if  the  principle  is  true  for 
polynomials  of  n  terms,  it  is  true  for  polynomials  of  n  +  1  terms. 

By  actual  multiplication  the  principle  has  been  proved  true  for 
polynomials  of  two,  three,  and  four  terms,  respectively.  There- 
fore, being  true  for  polynomials  of  four  terms,  the  principle  holds 
true  for  polynomials  of  five  terms  \  being  true  for  polynomials  of 


MATHEMATICAL   INDUCTION  407 

live  terms,  it  holds  true  for  polynomials  of  six  terms ;  and  so  on 
indefinitely. 

Hence,  the  principle  is  universally  true. 

4.   Let  it  be  required,  to  prove  by  mathematical  induction  that 
12  +  22  +  3^  4-  ...  +v?  =  i n(n  +  1) (2n  +  1).  (1) 

Supposing  that  (1)  is  true,  then,  Ax.  2, 

124.22+32+  ...  +/iH.(w+l)'=f<'*  +  IK2«+l)4-(n  +  l)0 
§  85,  =  \{n  -H  l)[»i(2  %  +  1)  +  6(n  +  1)] 

=  i(H  +  l)(2n2  +  7n+6) 
=  i(w  +  l)(n  +  2)(2H+3) 
§§56,85,  =K«+1)(^^+1)(2*"^+1)-     (2) 

By  comparing  (2)  with  (1)  it  is  seen  that  the  sum  of  the 
squares  of  the  first  {n  +  1)  integers  has  the  same  form  with 
respect  to  {n  + 1)  that  the  sum  of  the  squares  of  the  first  n  in- 
tegers has  with  respect  to  n ;  that  is,  it  has  been  proved  that  if 
the  formula  is  true  for  n  terms,  it  is  true  for  n  +  1  terms. 

It  can  be  verified  by  actual  trial  that  the  formula  is  true  for 
1,  2,  3,  4,  ...  terms,  as  far  as  we  please.  Supposing  that  the 
verification  stops  with  w  =  5.  By  the  above  proof,  since  the 
formula  is  true  for  five  terms  it  holds  true  for  six  terms  ;  being 
true  for  six  terms  it  holds  true  for  seven  terms ;  and  so  on  indefi- 
nitely. Hence,  the  principle  expressed  by  (1)  is  true  for  all 
integral  values  of  n. 

Examples 
Prove  by  mathematical  induction  that : 

1.  12  +  32  +  52+  ...  to«terms=-(2n-l)(2n  +  l). 

o 

2.  1  .  3  +  2  .  4  +  3  .  5  +  ...  to  w  terms  = -(2  ^2  +  9  w  +  7). 

6 

3.  1.3-f3.5-f5.7-f...tow  terms  =  -  (4  ^2  +  6  w  -  1). 

o 

4.  a  -f  ar  -\-  ai^  +  •••  to  w  terms  =    \ '-• 

1  —  r 

5.  ic2"  —  y2n  is  divisible  hj  x  +  y,  n  being  a  positive  integer. 


THE   BINOMIAL   THEOREM 


450.  The  Binomial  Theorem  derives  a  lormula  by  means  of 
which  any  power  of  a  binomial  may  be  expanded  into  a  series, 
whether  the  index  of  the  power  is  positive  or  negative,  integral 
or  fractional. 


POSITIVE  INTEGRAL  EXPONENTS 
451.   The  powers  of  (a  +  x),  expanded  in  §  221,  may  be  written 

2  •  1 

(a  +  x)-  =  a-  4-  2  ax  +  ^ — -  a^. 


(a  +  xy  =  a'  +  3a'x  +  'p-^ax^  +  - 
^  1-2  1 


(a  +  xy  =  a''  4-  5  a*x  +  ^^  aV  +  J 


5.4 
1.2' 


■of. 


3  1.2.3-4 

3-2.1 


3-4.5 


■a^. 


If  the  law  of  development  revealed  in  the  above  is  assumed  to 
apply  to  the  expansion  of  any  power  of  any  binomial,  as  the  wth 
power  of  (a  +  x),  the  result  is 

^  1-2  1-2-3  ^  ^ 

From  formula  (1)  it  is  evident  that  in  any  term, 

1.  The  exponent  of  x  is  1  less  than  the  number  of  the  term. 
Hence,  the  exponent  of  x  in  the  (r  +  l)th  term  is  r. 

2.  The  exponent  of  a  is  n  minus  the  exponent  of  x. 
Hence,  the  exponent  of  a  in  the  (r  +  l)th  term  is  w  —  r. 

408 


BINOMIAL    THEOREM  409 

3.  The  namber  of  factors  in  the  numerator  and  in  the  denomi- 
nator of  the  coefficient  is  1  less  than  the  number  of  the  term. 

Hence,  the  coefficient  of  the  (?*  -f  l)th  term  has  r  factors  in  the 
numerator  and  r  factors  in  the  denominator. 

Therefore,  the  (r  +  l)th,  or  general  term,  is 

n(n~l){n-2)--  tor  factors  ^„_,^^  ^2) 

1  •  2  •  3  •  •  •  to  >•  factors 

Since,  when  there  are  two  factors  in  the  numerator,  the  last  is 
w  —  1,  when  there  are  three  factors,  n  —  2,  when  there  are  four 
factors,  n  —  3,  etc. ;  when  there  are  r  factors,  the  last  is  n—  (r— 1), 
or  w  —  r  +  1.     Hence,  (2)  may  be  written 

n{n-V){n-  2)  -"  jn-r  +  l)  ,^. 

Hence,  the  full  form  of  (1)  is 

This  is  called  the  Binomial  Formula. 

452.  Since  it  has  already  been  proved,  by  actual  multiplica- 
tion, that  the  binomial  formula  is  true  for  the  second,  third, 
fourth,  and  fifth  powers  of  a  binomial,  it  remains  to  discover 
Avhether  it  is  true  for  powers  higher  than  the  fifth. 

If  the  binomial  theorem,  when  assumed  to  be  true  for  the  nth 
power,  can  be  j)roved  to  be  true  for  the  {n  +  l)th  power,  it  will 
then  have  been  proved  to  be  true  for  the  sixth  power,  since  it  is 
known  to  be  true  for  the  fifth  power ;  also  for  the  seventh  power, 
being  true  for  the  sixth  power ;  and  in  like  manner  for  each  suc- 
ceeding power. 

It  then  remains  to  prove  that  if  (I)  is  true  for  the  rtth  power, 
it  will  hold  true  for  the  (n  -f  l)th  power. 

To  find  the  expansion  of  (a  +  x)"+*  (I)  may  be  multiplied  by 
a-faj.  But  since  the  (r-f  l)th  term  of  the  product  will  be  the 
algebraic  sum  of  a  times  the  (r4-l)th  term  of  {a+xy  and  x  times 
the  rth  term  of  (a+x)",  (I)  should  be  prepared  for  multiplication 


410 


BINOMIAL   THEOREM 


by  writing  also  the  Hh  term,  obtained  from  the  (r  +  l)th  term  byj 
substituting  r  for  r  -f  1,  or  r  —  1  for  r.     Then  (I)  is  written 

(a  +  xy  =  a"  +  na^-'x  +  ^^^"^^a"  V  +  ... 

n{n-l){n-2)"'{n-r  +  2,)        +^      ^ 
■^  1.2.3."(r-l) 

^  n(n-l)(n-2).-.(n-9-  +  2)(n-r  +  l)^n-r^  ,         ,  ^n 
1  -2  -3  •••  (r  —  l)r 

Multiplying  both  members  by  a  +  x, 


(a  +  a;)"+i  =  a"+'  +  n 
+  1 


a"a;  + 


n  (yi  —  1) 


1-2 

+  n 


a"-'^x^  + 


+ 


n (n  —  1)  (w  —  2)  •••  (n  —  r  +  2)  (n  —  r  + 1) 


+ 


1.2.3-(r-l)r 

n(n  —  l)(n  —  2)  •••  (n  —  r  +  2) 


1.2.3..- (r-  1) 


,n+l 


.n+1 


=  a"+^  +  (n  +  l)a"a;  +  [V^^]_J^  + 

Ai-r+l       \A^(n-l)(n-2).-.(n-r  +  2)\  „_,^i        ^_^ 
"^V       r       "^  A  l-2.3...(r-l)  ; 

=  a»+i  +  (ri  + 1)  a»a;  +  ^^^-tl)^  a"  ia;2  +  . . . 

J.    *   ^ 

+(^)("^''  - 1^.^;  .-3^!.  (;^'  1-) ""  ^  '>-'"-^+ ■■■+' 

That  is, 

(a  +  «)«+i  =  a"+i  +  (n  +  1)  a"a;  +  (^^-±l)l^a»-V  +  ... 

(r.-fl)n(n-l)-.(n-r  +  2)^„_,^t^      ...  +a;»+i.    (II) 

Since  upon  comparison  it  may  be  seen  that  (II)  and  (I)  have 
the  same  form,  ri  + 1  in  one  taking  the  place  of  n  in  the  other, 
(II)  and  (I)  must  express  the  same  law  of  formation. 

Therefore,  if  the  formula  is  true  for  the  wth  power,  it  holds 
true  for  the  (n  +  l)th  power. 


BINOMIAL    THEOREM  411 

Hence,  the  binomial  formula  is  true  for  any  positive  integral 
exponent. 

This  method  of  proof  is  a  proof  by  Mathematical  Induction. 

463.  If  —  a;  is  substituted  for  x  in  (I),  the  terms  that  contain 
the  odd  powers  of  —  x  will  be  negative,  and  the  terms  that  con- 
tain the  even  powers  will  be  positive.     Therefore, 

(a-xY  =  a"  -  wa'-^x  +  ^<^  -^l^ju-i^s _  gH) 

1    ■   a 

If  a  =  1,  (I)  becomes 

454.  From  (I)  it  is  seen  that  the  last  factor  in  the  numerator 
of  the  coefficient  is  n  for  the  2d  term,  n  —  1  for  the  3d  term, 
n  —  2  for  the  4th  term,  n  —  (n  —  2),  or  2,  for  the  rith  term,  and 
n  —  {n  —  1),  or  1,  for  the  {n  +  l)th  term ;  and  that  the  coefficient 
of  the  (w  +  2)th  term,  and  that  of  each  succeeding  term,  contains 
the  factor  n  —  n,  or  0,  and  therefore  reduces  to  0.     Hence, 

Wlien  n  is  a  positive  integer,  the  series  formed  by  expanding 
{a  +  x)"  is  finite  and  has  w  -f  1  terms. 

455.  By  formula  (I),  when  n  is  a  positive  integer, 

/     .     \n        n  ,       «-i     ,  n(n  —  l)   -_o'i  ,         ,  n(w  —  1)  •••  2  •  1  _ 

(a  +  X)"  =  a"  +  «a"  'a;  +  -^^ ^a"  -or  +  •••  -f  -'^ ^. cc". 

1-2  1  •  2  •••  («  —  1)71 

/         .         \n  ^     >  n-1  ,     Mn  —  1)      „_2    •>     ,  .     n(n  —  1)    •••  2    •    1      _ 

(x  4-  a)"  =  af  -f  nx^^a  +  -^^ ^aj"  V  +  •••  +  -^^ ^- a", 

1-2  1  •  2  •••  (n  —  V)n 

A  comparison  of  the  two  series  shows  that : 
.  The  coefficients  of  the  latter  half  of  the  expansion  of  (a  +  a;)",  when 
n  is  a  positive  integer,  are  the  same  as  those  of  the  first  half,  written 
in  the  reverse  order. 

Examples 
1.   Expand  (3  a  -  2  by. 

Solution.  —  Substituting  3  a  for  a,  2  6  for  x,  and  4  for  n  in  (III), 
(3a-26)*  =  (3a)*-4(3«)3(26)+i^(.3a)2(26)2-l^|-l|(3a)(26)« 

1.2.3.4^      ^ 
=  81  a*  -  216  a%  +  216  a%^  -  96  afts  +  16  6*. 


412  BINOMIAL    THEOREM 

2.   Expand  f^  +  bx\\ 


Solution 


Since  (f  +  ''-y=[^l+2x)J=|(l  +  2.)S, 

(1  +  2  a;)5  may  be  expanded  by  (IV),  and  the  result  multiplied  by  -• 

0J4 


3. 
4. 

cpand : 

(b  -  7iy. 

(i+a-y. 

10. 

MJ- 

15. 

{''-:-j- 

5. 

(2-3aj)«. 

11. 

fa  _  x\\ 
\x     a) 

16. 

n~l           1 

(^x~^-xy. 

6. 

(x^  -  xf. 

\^    yJ 

17. 

{ax  "^  —b^xf. 

7. 

(x  +  x-y. 

12. 

18. 

f^~a      VbY 
\^/b      Va'J 

8 

(2  a  +  V^)^. 

13. 

(\/a^  +  </F«)3. 

1 —                    / — 

9 

(a  +  aVa)*. 

14. 

(2V2-^3f. 

19. 

m^u 

456.    To  find  any  term. 

Any  term  of  the  expansion  of  a  power  of  a  binomial  may  be 
obtained  by  substitution  in  (2)  or  (3),  §  451. 

In  the  expansion  of  a  power  of  the  difference  of  two  numbers,  as  (a  —  x)", 
since  the  exponent  of  x  in  the  (r  +  l)th  term  is  r,  the  sign  of  the  general 
term  is  +  if  ?•  is  even,  and  —  if  r  is  odd. 

Examples 
1.    Find  the  12th  term  of  (a  -  6)". 

Solution 

12th  term  =  14  •  13  •  12  .  11  ■  10  ■  0  ■  8  •  7  ■  6  .  5  ■  4    3 

1.2.3.4.5.6.7.8.9.10.11     ^     ^\i, 

=  _  14  •  13  -12  ^3511  =  _  364  a%n_         U^ 
Or 

By  §  4.55,  since  there  are  15  terms,  the  coefBcient  of  the  12th  terra,  or  the 
4th  term  from  the  end,  is  equal  to  that  of  the  4th  term  from  the  beginning. 

.-.  12th  term  =  -  ^jJ^jJ3  oSju  _  _  364  38511, 
1  •  ^  •  o 


BINOMIAL    THEOREM  413 

2.  In  the  expansion  of  (ic^+2iB)",  find  the  term  containing  x^. 
Solution  .—  Since  (x^  +  2  x)"  =  fa^^f  1  +  -  )  1    =  a;''^(  1  +  -  )    ,  every  term 

of  the  series  expanded  from  M  +  -  j     will  be  multiplied  by  x"^^^. 

Hence,  the  term  sought  is  that  which  contains  |  -  J  ,  or  —  ;  that  is,  the 
(7  +  l)th,  or  8th  term.  ^^'  ^' 

8th  term  =  x'^'^  W-^^-^-'^  f-V  =  42240  xi*. 

1.2.3.4  vx; 

3.  Find  the  4th  term  of  (a  +  2)^". 

4.  Find  the  4th  term  of  {x  -  3?/)'2. 

5.  Find  the  8th  term  of  {x  +  yf. 

6.  Find  the  5th  term  of  {x-1  yf\ 
1.    Find  the  3d  term  of  {a-  -  a-y. 

8.  Find  the  20th  term  of  (1  +  xf\ 

9.  Find  the  16th  term  of  (1-2  x)"^. 

10.  Find  the  middle  term  of  {a  +  3  h)\ 

11.  Find  the  6th  term  of  (x-\-^°- 


12.    Find  the  middle  term  off ^  —  -]  • 

13.    Find  the  two  middle  terms  of  f  -  - 

aj 

lA        ^^'^n^^    fVio  r>r«oflRr>ionf   r>-f  n^  in   i-\\o  a•v^ 

riancinn    r\f  1 

n^  _1_  «^5 

457.  The  formula  given  for  the  expansion  of  (a  +  x)"  is  true, 
tinder  certain  conditions,  for  all  commensurable  values  of  «, 
whether  they  are  positive,  negative,  integral,  or  fractional,  and 
the  student  will,  therefore,  be  able  to  expand  such  expressions ; 
but  the  proof  for  negative  and  fractional  exponents  and  the  dis- 
cussion of  the  conditions  under  which  the  expansion  for  these 
exponents  gives  the  true  valvie  of  (a  +  aj)"  will  be  deferred.  (See 
pages  520-523.) 

In  the  expansion  of  (o  +  x)",  if  n  is  negative  or  fractional, 

none  of  the  binomial  coefficients,     ^^  ~ — L  — — ~    >\'  ~  -^i  g^p 

1 • 2     '  1-2.3 

can  become  0 ;  consequently,  when  such  exponents  are  given,  the 
series  developed  can  have  no  end. 


414  BINOMIAL    THEOREM 

Examples 

1.  Expand  (1  —  y)-'^  and  find  its  (r  +  l)th  term. 
Solution.  —  Substituting  1  for  a,  y  for  x,  and  —  1  for  n  in  (III), 

(1  -  2/)-l  =  1-1  -(-  1)  l-2y  +  ^11=^  1-3  y2  _  ^iI:i|lIzLlll-4y8  +  ... 

=  1  +  2/  + 2/- -I- 2/^  +  •••• 
Tlie  (r  +  l)th  term  is  evidently  y^. 

Since  (1  —  y)"^  = ,  the  above  expansion  of  (1  —  j/)"^  may  be  verified 

by  division.  ~  ^ 

2.  Expand  (a  -|-  x)^  to  five  terms  and  find  the  10th  term. 

Expand  to  four  terms : 

3.  {a  +  h)^.  8.  VIT^.  13.  (l+a;)l 

4.  (a-|-6)"i  9.  V(9  -  xf.  14.  (l-ha)-\ 

5.  (a -6)1  10.  {a^-x-^)i.  15.  (1  _  a)-i. 

6.  </{a  -  &)•''.  11.    (a^-x*)-«.  16.    (1  -  a;)-^ 

1 


7.      ,^  12.    (     ,-       3/-)-  17.    (l-xV 

^(a-6/  VVa-Va;/  ^  ^ 

18.  Find  the  (r  +  l)th  term  of  {a  +  x)^. 

19.  Show  that 

{\-x-x^-'  =  1  +  0;  +  2.t2  +  3ar'+  5a;*  +  8ar'+  13.T«-1-  21  x" + 

20.  Eind  the  square  root  of  24  to  three  decimal  places. 
Solution.     V24  =(24)2  =  (25  -  1)^=  (25)^(1  -  ^.)^=  5(1  -  ^V)^- 

=  5  -  .1  -  .001  -  .00002 =  4.89898  -  =  4.899,  nearly. 

Find  the  values  of  the  following  to  three  decimal  places : 

21.  Vs.  23.     V26.  25.    V9. 

22.  VT7.  24.    -s/25.  26.    </M. 


LOGARITHMS 


458.  1.   What  power  of  3  is  9  ?    27?    81?    243?    729? 

2.  What  power  of  5  is  25  ?    125?    625?    3125?    5?    1?    i? 

3.  Express  100  as  a  power  of  10 ;  1000  as  a  power  of  10 ;  10,000 
as  a  power  of  10 ;  10  as  a  power  of  10 ;  1  as  a  power  of  10. 

459.  The  exponent  of  the  power  to  which  a  fixed  number  called 
the  Base  must  be  raised  in  order  to  produce  a  given  number  is 
called  the  Logarithm  of  the  given  number. 

When  10  is  the  base,  the  logarithm  of  100  is  2,  for  100  =  10^  ;  the  logarithm 
of  1000  is  3,  for  1000  =  10^ ;  the  logarithm  of  10,000  is  4,  for  10,000  =  10*. 

460.  When  a  is  the  base,  x  the  exponent,  and  m  the  given 
number,  x  is  the  logarithm  of  the  number  m  to  the  base  a. 

It  is  written  log„  m  =  x. 

When  the  base  is  10,  it  is  not  indicated. 

Thus,  the  logarithm  of  100  to  the  base  10  is  2.    It  is  written  log  100  =  2. 

461.  Logarithms  may  be  computed  with  any  arithmetical 
number  except  unity  as  a  base,  but  the  base  of  the  Common  or 
Briggs  System  of  logarithms  is  10. 

Since  10°    =  1,  the  logarithm  of  1  is  0. 
Since  10'    =  10,  the  logarithm  of  10  is  1. 
Since  10^    =  100,  the  logarithm  of  100  is  2. 
Since  10^    =  1000,  the  logarithm  of  1000  is  3. 
Since  10~'  =  ^,  the  logarithm  of  .1  is  —  1. 
Since  10~^  =  y^^,  the  logarithm  of  .01  is  —  2. 

462.  It  is  evident,  then,  that  the  logarithm  of  any  number 
between  1  and  10  is  a  number  greater  than  0  and  less  than  1. 
For  example,  the  logarithm  of  4  is  approximately  0.6021.  Again, 
the  logarithm  of  any  number  between  10  and  100  is  a  number 
greater  than  1  and  less  than  2.  For  example,  the  logarithm  of 
50  is  approximately  1.6990. 

416 


416  ^  LOGARITHMS 

Most  logarithms  are  incommensurable  uurabers.  All  the  laws 
established  for  commensurable  exponents  apply  also  to  incom- 
mensurable exponents.  The  proofs  for  incommensurable  expo- 
nents are  given  in  §  436. 

463.  The  integral  part  of  a  logarithm  is  called  the  Characteris- 
tic; the  fractional  or  decimal  part,  the  Mantissa. 

In  log  60  =  1.6990,  the  characteristic  is  1  and  the  niarcissa  .6990. 

464.  The  following  examples  will  illustrate  the  cJiaracteristic 
and  mantissa,  and  their  significance : 

log  458/   =  3.6609 ;  that  is,  4580  =  lO'^'^. 

that  is,  458.0  =  lO^*^. 

that  is,  45.80  =  10^*^. 

that  is,  4.580  =10"'^. 

that  is,  .4580  =  10-i+«*». 

that  is,  .0458  ^  10-2+ ««»; 
that  is,  .00458  =  10-''+««». 


log458A  =2.6609 
log45.8i  =1.6609 
log4.58J>  =0.6609 
log  .4580  =1.6609 
log  .0458  =  2.6609 
log  .00458  =  3.6609 


From  the  above  examples  it  is  evident  that : 

465.  Principles.  —  1.  The  characteristic  of  the  logarithm  of  a 
mimber  greater  than  1  is  positive  and  1  less  than  the  number  of 
digits  in  its  integral  part.  i 

2.  The  characteristic  of  the  logarithm  of  a  decimal  is  negative  and 
numerically  1  greater  than  the  number  of  ciphers  immediately  follow- 
ing the  decimal  point. 

466.  To  avoid  writing  a  negative  characteristic  before  a  positive 
mantissa,  it  is  customary  to  add  10  or  some  multiple  of  10  to  the 
negative  characteristic,  and  to  indicate  that  the  number  added  is 
to  be  subtracted  from  the  whole  logarithm. 

Thus,  f.6609  is  written_9.6609  -  10  ;  2.3010  is  written  8.3010  -  10  ;  14.9031 
is  written  6.9031  -  20  ;  28.8062  is  written  2.8062  -  30  ;  etc. 

467.  It  is  evident,  also,  from  the  examples,  that  in  the  loga- 
rithms of  numbers  expressed  by  the  same  figures  in  the  same 
order,  the  decimal  parts,  or  mantissas,  are  the  same,  and  that  the 
logarithms  differ  only  in  their  characteristics.  Hence,  tables  of 
logarithms  contain  only  the  mantissas. 


LOGARITHMS  417 

468.  The  table  of  logarithms  on  the  two  following  pages  gives 
the  decimal  parts,  or  mantissas,  correct  to  four  places,  for  the 
common  logarithms  of  all  numbers  from  1  to  1000. 

469.  To  find  the  logarithm  ofjajumfeer.  - ' 

Examples 

1.  Find  the  logarithm  of  765.  >/ 

Solution.  —  In  the  following  table  the  letter  N  designates  a  vertical 
column  of  numbers  from  10  to  99  inclusive,  and  also  a  horizontal  row  of 
figures  0,  1,  2,  3,  4,  5,  6,  7,  8,  9.  The  first  two  figures  of  765  appear  as  the 
number  76  in  the  vertical  colunm  marked  N  on  page  419,  and  the  third  figure 
5  in  the  horizontal  row  marked  N. 

In  the  same  horizontal  row  as  76  are  found  the  mantissas  of  the  logarithms 
of  the  numbers  760,  761,  762,  763,  764,  765,  etc.  The  mantissa  of  the  loga- 
rithm of  765  is  found  in  this  row  under  5,  the  third  figure  of  765.  It  is  8837 
and  means  .8837. 

By  Prin.  1,  the  characteristic  of  the  logarithm  of  765  is  2. 

Hence,  the  logarithm  of  765  is  2.8837. 

2.  Find  the  logarithm  of  4.  ^^^ 

SoLiTiox.  — Although  the  numbers  in  the  table  ap^^^B  begin  with  100, 
the  table  really  includes  all  numbers  from  1  to  1000,  siiSBKmbers  expressed 
by  less  than  three  figures  may  be  expressed  by  three  figures  by  adding  deci- 
mal ciphers.  Since  4  =  4.00,  and  since,  §  467,  the  mantissa  of  the  logarithm 
of  4.00  is  the  same  as  that  of  400,  which  is  .6021,  the  mantissa  of  the  loga- 
rithm of  4  is  .6021. 

By  Prin.  1,  the  characteristic  of  the  logarithm  of  4  is  0. 

Therefore,  the  logarithm  of  4  is  0.6021. 

Verify  the  following  from  the  table  : 

3.  log  10    =L0000.  9.  log  .2      =9.8010-10. 

4.  log  100  =  2.0000.  10.  log  542  =2.7340. 

5.  log  110  =  2.0414.  11.  log  345  =2.5378. 

6.  log  2      =0.3010.  12.  log  5.07  =  0.7050. 

7.  log  20    =1.3010.  13.  log  78.5  =  1.8949. 

8.  log  200  =  2.3010.  14.  log  .981  =  9.9917  -  10. 

ADV.    Al.O.—  27 


418 


LOGARITHMS 


Table  of  Common  Logarithms 


N 

0 

1 

2 

3 

4 

5 

6 

7 

8 

9 

lO 

0000 

0043 

0086 

0128 

0170 

0212 

0253 

0294 

0334 

0374 

II 

0414 

0453 

0492 

0531 

0569 

0607 

0645 

0682 

0719 

0755 

12 

0792 

0828 

0864 

0899 

0934 

0969 

1004 

1038 

1072 

1106 

13 

"39 

"73 

1206 

1235 

1271 

1303 

1335 

1367 

1399 

1430 

14 

1461 

1492 

1523 

i£^3 

1584 

1614 

1644 

1673 

1703 

1732 

15 

1761 

1790 

1818 

184^ 

1875 

1903 

1931 

1959 

1987 

2014 

16 

2041 

2068 

2095 

2122 

2148 

2175 

2201 

2227 

2253 

2279 

17 

2304 

2330 

2355 

2380 

2405 

2430 

2455 

2480 

2504 

2529 

18 

2553 

2577 

2601 

2625 

2648 

2672 

2695 

2718 

2742 

2765 

19 

2788 

2810 

2833 

2856 

2878 

2900 

2923 

2945 

2967 

2989 

20 

3010 

3032 

3054 

3075 

3096 

3118 

3139 

3160 

3181 

3201 

21 

3222 

3243 

3263 

3284 

3304 

3324 

3345 

3365 

3385 

3404 

22 

3424 

3444 

3464 

3483 

3502 

3522 

3541 

3560 

3579 

3598 

23 

3617 

3636 

3655 

3674 

3692 

37" 

3729 

3747 

3766 

3784 

24 

3802 

3820 

3838 

3856 

3874 

3892 

3927 

3945 

3962 

25 

3979 

3997 

4014 

4031 

4048 

4065 

4082 

4099 

4116 

4133 

26 

4150 

4166 

4183 

4200 

4216 

4232 

4249 

4265 

4281 

4298 

27 

4314 

4330 

4346 

4362 

4378 

4393 

4409 

4425 

4440 

4456 

28 

4472 

4487 

4502 

4518 

4533 

4548 

4564 

4579 

4594 

4609 

29 

4624 

4639 

4654 

4669 

4683 

4698 

4713 

4728 

4742 

4757 

30 

4771 

4786 

4800 

4814 

4829 

4843 

4857 

4871 

4886 

4900 

31 

4914 

4928 

4942 

4955 

4969 

4983 

4997 

501 1 

5024 

5038 

32 

5051 

5065 

5079 

5092 

5105 

5"9 

5132 

5145 

5159 

5172 

33 

5185 

5198 

5211 

5224 

5237 

5250 

5263 

5276 

5289 

5302 

34 

5315 

5328 

5340 

5353 

5366 

5378 

5391 

5403 

5416 

5428 

35 

5441 

5453 

5465 

5478 

5490 

5502 

S5H 

5527 

5539 

5551 

36 

55^3 

5575 

5587 

5599 

5611 

5623 

5635 

5647 

5658 

5670 

37 

5682 

5694 

5705 

5717 

5729 

5740 

5752 

5763 

5775 

5786 

38 

5798 

5809 

5821 

5832 

5843 

5855 

5866 

5877 

5888 

5899 

39 

5911 

5922 

5933 

5944 

5955 

5966 

5977 

5988 

5999 

6010 

40 

6021 

6031 

6042 

6053 

6064 

6075 

6085 

6096 

6107 

6117 

41 

6128 

6138 

6149 

6160 

6170 

6180 

61^1 

6201 

6212 

6222 

42 

6232 

6243 

6253 

6263 

6274 

6284 

6294 

6304 

6314 

6325 

43 

6335 

6345 

6355 

6365 

6375 

6385 

6393 

6405 

6415 

6425 

44 

6435 

6444 

6454 

6464 

6474 

64041. 

6493 

6503 

6513 

6522 

45 

6532 

6542 

6551 

6561 

6571 

6580 

6590 

6599 

6605 

6618 

46 

6628 

6637 

6646 

6656 

6665 

6675 

6684 

6693 

6702  "6712  1 

47 

6721 

6730 

673^1 

674t> 

6758 

6767 

6776 

6785 

6794 

6803 

48 

6812 

6821 

6830 

6839 

6848 

6857 

6866 

6875 

6884 

6893 

49 

6902 

6911 

6920 

6928 

6937 

6946 

6955 

6964 

6972 

6981 

50 

6990 

6998 

7007 

7016 

7024 

7033 

7042 

7050 

7059 

'067 

51 

7076 

7084 

7093 

7101 

7110 

7118 

7126 

7>35 

7 '43 

7152 

52 

7160 

7168 

7177 

7185 

7193 

7202 

7210 

TJlS 

•  7326 

J^ZS 

53 

7243 

7251 

7259 

7267 

7275 

7284 

7292 

') 

54 

7324 

7332 

7340 

7348 

7356 

7364 

7372 

7 

r®" 

9 

N 

0 

1 

2 

3 

4 

~5~ 

6 

y 


'y 


.  r 


V 


LOGARITHMS 


419 


Table  of  Common  Logarithms 


N 

0 

1 

2 

0 

3 

4 

5 

6 

7 

8 

9 

55 

7404 

7412 

7419 

7427 

7435 

7443 

7451 

7459 

7466 

7474 

56 

7482 

7490 

7497 

7505 

7513 

7520 

7528 

7536 

7543 

7551 

57 

7559 

7566 

7574 

7582 

7589 

7597 

7604 

7612 

7619 

7627 

58 

7634 

7642 

7649 

7657 

7664 

7672 

7679 

7686 

7694 

7701 

59 

7709 

7716 

7723 

7731 

7738 

7745 

7752 

7760 

7767 

7774 

60 

7782 

7789 

7796 

7803 

7810 

7818 

7825 

7832 

7839 

7846 

61 

7853 

7860 

7868 

7875 

7882 

7889 

7896 

7903 

7910 

7917 

62 

7924 

7931 

7938 

7945 

7952 

7959 

7966 

7973 

7980 

7987 

63 

7993 

8000 

8007 

8014 

8021 

8028 

8035 

8041 

8048 

8055 

64 

8062 

8069 

8075 

8082 

8089 

8096 

8102 

8109 

8116 

8122 

65 

8129 

8136 

8142 

8149 

8156 

8162 

8169 

8176 

8182 

8189 

66 

8195 

8202 

8209 

8215 

8222 

8228 

8235 

8241 

8248 

8254 

67 

8261 

8267 

8274 

8280 

8287 

8293 

8299 

8306 

8312 

8319 

68 

8325 

8331 

8338 

8344 

8351 

8357 

8363 

8370 

8376 

8382 

69 

8388 

8395 

8401 

8407 

8414 

8420 

8426 

84^ 

8439 

8445 

70 

8451 

8457 

8463 

8470 

8476 

8482 

8488 

8494 

8500 

8506 

7X 

8513 

8519 

8525 

8531 

8537 

8543 

8549 

8555 

8561 

8567 

72 

8573 

8579 

8585 

8591 

8597 

8603 

8609 

8615 

8621 

8627 

73 

8633 

8639 

8645 

8651 

8657 

8663 

8669 

8675 

8681 

8686 

74 

8692 

8698 

8704 

8710 

8716 

8722 

8727 

8733 

8739 

8745 

75 

8751 

8756 

8762 

8168 
8fc5 

8774 

8779 

8785 

8791 

8797 

8802 

76 

8808. 

8814 

8820 

8831 

-8837 

8842 

8848 

8854 

8859 

77 

8865 

8871 

8876 

8882 

8887 

8893 

8899 

8904 

8910 

8915 

78 

8921 

8927 

8932 

8938 

8943 

8949 

8954 

8960 

8965 

8971 

79 

8976 

8982 

8987 

8993 

8998 

9004 

9009 

9015 

9020 

9025 

80 

9031 

9036 

9042 

9047 

9053 

9058 

9063 

9069 

9074 

9079 

81 

9085 

9090 

9096 

9101 

9106 

9112 

9117 

9122 

9128 

9133 

82 

9138 

9143 

9149 

9154 

9159 

9165 

9170 

9175 

9180 

9186 

83 

9191 

9196 

9201 

9206 

9212 

9217 

9222 

9227 

9232 

9238 

84 

9243 

9248 

9253 

9258 

9263 

9269 

9274 

9279 

9284 

9289 

85 

9294 

9299 

9304 

9309 

9315 

9320 

9325 

9330 

9335 

9340 

86 

9345 

9350 

9355 

9360 

9365 

9370 

9375 

9380 

9385 

9390 

87 

9395 

9400 

9405 

9410 

9415 

942e. 

9425 

9430 

9435 

9440 

88 

9445 

9450 

9455 

9460 

9465 

9469 

9474 

9479 

9484 

9489 

89 

9494 

9499 

9504 

9509 

9513 

9518 

9523 

9528 

9533 

9538 

90 

.9  = 

i?ll^ 

9557 

9562 

9566 

9571 

9576 

9581 

9586 

91' 

c 

9605 

9609 

9614 

9619 

9624 

9628 

9633 

92 

9 

!  9647 ' 

9652 

9657 

9661 

9666 

9671 

9675 

9680 

93 

c 

9601 

9699 

9703 

9708 

9713 

9717 

9722 

9727 

94   < 

97J5 

gjy> 

9754 

9759 

9763 

9768 

9773 

95 

9791 

9795 

9800 

9805 

9809 

9814 

9818 

96 

:      9836 

9841 

9845 

9850 

9854 

9859 

9863 

97 

7   9^1 

9886 

9890 

9894 

9899 

9903 

9908 

98 

I    i  9926 

99*30 

9934 

9939 

9943 

9948 

9952 

99 

N|  0 

1 

'5 

9969 

9974 

9978 

9983 

9987 

9991 

9996 

f 

3 

4 

5 

6 

7 

8 

9 

1 

420  LOGARITHMS 

15.  Find  the  logarithm  of  6253. 

Solution.  —  Since  the  table  contains  the  mantissas  not  only  of  the  loga- 
rithms of  numbers  expressed  by  three  figures,  but  also  of  logarithms  expressed 
by  four  figures  when  the  last  figure  is  0,  the  mantissa  of  the  logarithm  of  625  is 
first  found,  since,  §  467,  it  is  also  the  mantissa  of  the  logarithm  of  6250.  It 
is  found  to  be  .7959. 

The  next  greater  mantissa  found  in  the  table  is  .7966,  which  is  the  man- 
tissa of  the  logarithm  of  626  or  of  6260.  Since  the  numbers  6250  and  6260 
differ  by  10,  and  the  mantissas  of  their  logarithms  differ  by  7  ten-thou- 
sandths, it  may  be  assumed  as  sufficiently  accurate  that  each  increase  of  1 
unit,  as  6250  increases  to  6260,  produces  a  corresponding  increase  of  .1  of  7 
ten-thousandths  in  the  mantissa  of  the  logarithm.  Consequently,  3  added 
to  6250  will  add  .3  of  7  ten-thousandths,  or  2  ten-thousandths,  to  the  man- 
tissa of  the  logarithm  of  6250  for  the  mantissa  of  the  logarithm  of  6253. 

Hence,  the  mantissa  of  the  logarithm  of  6253  is  .7959  -f  .0002,  or  .7961. 

Since  the  number  is  an  integer  expressed  by  4  digits,  the  characteristic  is 
3  (Prin.  1). 

Therefore,  the  logarithm  of  6253  is  3.7961. 

Note. — The  difference  between  two  successive  mantissas  in  the  table  is 
called  the  Tabular  Difference. 

Find  the  logarithm  of 

16.  1054.  20.  21.09.  24.  .09095. 

17.  1272.  21.  3.060.  25.  .10125. 

18.  .0165.  22.  441.1.  26.  54.675. 

19.  1906.  23.  .7854.  27.  .09885. 

470.    To  find  a  number  whose  logarithm  is  given. 


Examples 
1.    Find  the  number  whose  logarithm  is  0.9472. 

Solution. — The  two  mantissas  adjacent  to  the  given  mantissa  are  .9469 
and  .9474,  corresponding  to  the  numbers  8.85  and  8.86,  since  the  given 
characteristic  is  0.  The  given  mantissa  is  3  ten-thousandths  greater  than 
the  mantissa  of  the  logarithm  of  8.85,  and  the  mantissa  of  the  logarithm  of 
8.86  is  5  ten-thousandths  greater  than  the  mantissa  of  the  logarithm  of  8.85. 

Since  the  numbers  8.85  and  8.86  differ  by  1  one-hundredth,  and  the  man- 
tissas of  their  logarithms  differ  by  5  ten-thousandths,  it  may  be  assumed  as 
sufficiently  accurate  that  each  increase  of  1  ten-thousandth  in  the  mantissa 
is  produced  by  an  increase  of  |  of  1  one-hundredth  in  the  number.     Conse- 


LOGARITHMS  421 

quently,  an  increase  of  3  ten-thousandths  in  the  mantissa  is  produced  by  an 
increase  of  f  of  1  one-hundredth,  or  .00(3,  in  the  number. 
Hence,  tlie  number  whose  logarithm  is  0.9472  is  8.856. 

2.    Find  the  number  whose  logarithm  is  9.4180  — 10. 

Solution. — Given  mantissa,        .4180.- 

Mantissa  next  less,  .4166 ;  ^gures  corresponding,  261. 


Difference,  14 

Tabular  difference,  17)14(.8 

Hence,  the  figures  corresponding  to  the  given  mantissa  are  2618. 
Since  the  characteristic  is  9  —  10,  or  —J,  the  number  is  a  decimal  with  no 
ciphers  immediately  following  the  decimal  point. 

Hence,  the  number  whose  logarithm  is  9.4180  —  10  is  .2618. 

Find  the  number  corresponding  to.  '  ^    '      i 

3.  0.3010.                   8.   3.9545.  13.  9.3685-10.     ' 

4.  1.6021.                   9.   0.8794.  14.  8.9932-10. 
6.   2.9031.                 10.    2.9371.  15.  8.9535-10. 

6.  1.6669.  11.    0.8294.  16.   7.7168-10. 

7.  2.7971.  12.    1.9039.  17.   6.7016-10. 

471.  Multiplication  by  logarithms. 

•  Since  logarithms  are  the  exponents  of  the  powers  to  which  a 
constant  number  is  to  be  raised,  it  follows  that  : 

472.  Prixciple.  —  The  logarithm  of  the  product  of  two  or  more 
numbers  is  equal  to  the  sum  of  their  logarithms;  that  is, 

To  any  base,  log  (win)  =  log  m  -j-  log  n. 

The  above  principle  may  be  established  as  follows : 
Let  logo  m  =  X  and  logo  n  =  y,  a  being  any  base. 

It  is  to  be  proved  that     loga  (jwn)  =  x  +  y. 
§  459,  •  a'  =  w, 

and  *«»  =  n. 

Multiplying,  §§  240,  436,      '   0*+*  =  m». 
Hence,  §  460,  loga  (inn)  =  z  +  y 

=  log„  m  +  loga  n. 


422  LOGARITHMS 

Examples 
1.   Multiply  .0381  by  77. 

Solution 
Prin.,  log  (.0381  x  77)  =  log  .0381  +  log  77. 

log  .0381=  8.5809 -10 
log.  77  =  1.8865 
Sum  of  logs  =  10.4674  -  10 
=  0.4674 
0.4674  =  log  2.934. 
.-..0381  X  77  =  2.934.' 

Note.  —  Three  figures  of  a  number  corresponding  to  a  logarithm  may  be 
found  from  this  table  with  absolute  accuracy,  and  in  most  cases  the  fourth 
will  be  correct.  In  finding  the  logarithms  of  numbers  or  the  numbers  corre- 
sponding to  logarithms,  allowance  should  be  made  for  the  figures  after  the 
fourth,  whenever  they  express  .5  or  more. 

Multiply : 

2.  3.8  by  56.  6.  2.26  by  85.  10.  1.414  by  2.829. 

3.  72  by  39.  7.  7.25  by  240.  11.  42.37  by  .236. 

4.  8.5  by  6.2.  8.  3272  by  75.  12.  2912  by  .7281. 

5.  1.64  by  35.  9.  .892  by  .805.  13.  289  by  .7854. 

473.  Division  by  logarithms. 

Since  the  logarithms  of  two  numbers  to  a  common  base  repre- 
sent exponents  of  the  same  number,  it  follo.ws  that : 

474.  Principle.  —  Tlie  logarithm  of  the  quotient  of  two  numbers 
is  equal  to  the  logarithm  of  the  dividend  min^s  the  logarithm  of  the 
divisor;  that  is, 

To  any  base,       log  (m  -f-  w)  =  log  m  —  log  n. 

The  above  principle  may  be  established  as  follows: 

Let  loga  m  =  x  and  logo  n  =  y,  a  being  any  base. 

It  is  to  be  proved  that      loga(TO  -;-  n)  =  x  —  y. 
§  459,  a''  =  m  and  as  =  n. 

Dividing,  §§  248,  436,     a^-y  -m-i-n. 
Hence,  §  460,  loga(m  -=-  n)  =  a;  —  y 

=  logo  m  -  logo  n. 


LOGARITHMS  423 

Examples 

1.  Divide  .00468  by  75. 

Solution 

Prin.,  log  (.00468  --  75)  =  log  .00468  -  log  76. 

log  .00468  =  7.6702  -  10 

log  75         =  1.8751 

Difference  of  logs  =  5.7951  -  10 

5.7951  -  10  =  log  .00006239. 

.-.  .00468  -f-  75  =  .00006239. 

2.  Divide  12.4  by  16. 

Solution 

Prin.,  log  (12.4  -^  16)  =  log  12.4  -  log  16. 

log  12.4  =  1.0934  =  11.0934  -  10 

log  16     = 1.2041 

Difference  of  logs  =   9.8893  -  10 

9.8893  -  10  =  log  .775. 

.-.  12.4  --  16  =  .lib. 

Suggestion.  —  The  positive  part  of  the  logarithm  of  the  dividend  may 
be  made  to  exceed  that  of  the  divisor,  if  necessary,  by  adding  10  —  10  or 
20  -  20,  etc. 

Divide : 

3.  3025  by  55.     8.  10  by  3.14.      13.  1  by  40. 

4.  4096  by  32.     9.  .0911  by  .7854.   14.  1  by  75. 

'  5.  3249  by  57.    10.  2.816  by  22.5.    15.  200  by  .5236. 

6.  .2601  by  .68.    11.  4  by  .00521.     16.  300  by  17.32. 

7.  3950  by  .250.    12.  26  by  .06771.    17.  .220  by  .3183. 

475.   Extended  operations  in  multiplication  and  division. 

Since  dividing  by  a  number  is  equivalent  to  multiplying  by  its 
reciprocal,  for  every  operation  of  division  an  operation  of  multi- 
plication may  be  substituted.  In  extended  operations  in  multipli- 
cation and  division  with  the  aid  of  logarithms,  the  latter  method 
of  dividing  is  the  more  convenient. 


424  LOGARITHMS 

476.  The  logarithm  of  the  reciprocal  of  a  number  is  called  the 
Cologarithm  of  the  number. 

The  cologarithm  of  100  is  equal  to  the  logarithm  of  j^,  which  is  —  2. 
The  cologarithm  of  100  is  —  2  is  abbreviated  to  colog  100  =  —  2. 

477.  Since  the  logarithm  of  1  is  0  and  the  logarithm  of  a 
quotient  is  obtained  by  subtracting  the  logarithm  of  the  divisor 
from  that  of  the  dividend,  it  is  evident  that  the  cologarithm  of  a 
number  is  0  minus  the  logarithm  of  the  number,  or  the  logarithm 
of  the  number  with  the  sign  of  the  logarithm  changed ;  that  is,  if 
log„  m  =  x,  then,  colog„  m  =  —x. 

Since  subtracting  a  number  is  equivalent  to  adding  it  with  its 
sign  changed,  it  follows  that : 

478.  Principle.  —  Instead  of  subtracting  the  logarithm  of  the 
divisor  from  the  logarithm  of  the  dividend,  the  cologarithm  of  the 
divisor  may  be  added  to  the  logarithm  of  the  dividend;  that  is, 

To  any  base,      log  (m  ~n)  =  log  m  +  colog  n. 

Examples 

,     ^.    .  ,,         1        r  .063  X  58.5  X  799  ,     ,        .,, 

1.   Find  the  value  of  ^  ^^      ^  ^  ^ — —  by  logarithms. 

458  X  15.6  X  .029    "^     ^ 

Solution 
.068  X  58.5  X  799  ^  ^gg  ^  ,3  ,  ^  799  ^  J-  x  J-  x  ^. 
458  X  15.6  X  .029  458      15.6     .029 

log  .06.3=  8.7993-10 

log  58.5=  1.7672 

log  799  =  2.9025 

colog  458  =  7.3391  -  10 

colog  15.6=  8.8069-10 

colog  .029=  1.5376 


log  of  result  =  31.1526  -  30 
=    1.1526. 
.*.  result  =  14.21. 


LOGARITHMS  425 


Find  the  value  of 


110  X  3.1  X  .653  g   15  X  .37  x  26.16 

33  X  7.854  X  1.7* 
6000  X  5  X  29 


11  X  8  X  .18  X  6.67 

78 

x52x 

1605 

338 

X  767 

x431 

.5 

x.315 

x428 

.7854  X  25000  x  81.7 
^   3.516  X  485  X  65        ^ 

3.33  X  17  X  18  X  73*        *  .317  x  .973  x  43.7 

479.  Involution  by  logarithms. 

Since  logarithms  are  simply  exponents,  it  follows  that: 

480.  Principle.  —  Tlie  logarithm  of  a  poiver  of  a  number  is 
equal  to  the  logarithm  of  the  number  multiplied  by  the  index  of  the 
power ;  that  is, 

To  any  base,  log  m"  =  n  log  m. 

The  above  principle  may  be  established  as  follows : 

Let  loga  m  =  05,  and  let  n  be  any  number,  a  being  any  base. 

It  5s  to  be  proved  that  loga  m"  =  nx- 

§  459,  a^  =  m. 

Raising  each  member  to  the  nth  power, 

Ax.  6  and  §§  249,  436,  a*"  =  m». 

Hence,  §  460,  loga  w»  =  nx  =  n  log,  m. 

Examples 
1.   Find  the  value  of  .25^. 

Solution 
Prin,,  log  .252  =  2  log  .25. 

log  .26  =    9.3979  -  10. 
2  log  .25  =  18.7958  -  20 
=    8.7958  -  10. 
8.7958 -10  =  log  .06249. 
.-.  .252  =  .06249. 

Note. — By  actual  multiplication  it  is  found  that  .252  =  .0625,  whereas 
the  result  obtained  by  the  use  of  the  table  is  .06249. 

Also,  by  multiplication,  182  _  324^  whereas  by  the  use  of  the  table  it  is 
found  to  be  324.1.  Such  inaccuracies  must  be  expected  when  a  four-place 
table  is  used. 


426  LOGARITHMS 

Find  by  logarithms  the  value  of 

2.  V.  7.    .782  12.  4.073.  17.  {^)\ 

3.  11=*.  8.    8.051  13.  .5433.  18.  {\f. 

4.  47='.  9.    8.332.  14.  1\  19.  (tS^)*. 

5.  4.92.  10.   6.613.  15  1 02*.  20.  (A^/. 

6.  5.22.  11    JI42  16.  1.7383.  21.  {^\^)\ 

481.  Evolution  by  logarithms. 

Since  logarithms  are  simply  exponents,  it  follows  that: 

482.  Principle.  —  The  logarithm  of  the  root  of  a  number  is  equal 
to  the  logarithm  of  the  number  divided  by  the  index  of  the  require.-^ 
root;  that  is, 

To  any  base,  log  Vm  =  ^^S^- 

n 

The  above  principle  may  be  established  as  follows: 

Let  logo  m  =  X,  and  let  n  be  any  number,  a  being  any  base. 

It  is  to  be  proved  that  loga  y/m  =  x  ^  n. 

§  459,  a"^  =  m. 

Taking  the  ?ith  root  of  each  member, 

Ax.  7  and  §§  250,  436,  a^-"  =  Vm. 

Hence,  §  460,  log„  v^m  =  x  ^  n  =  \9SsJB. 

n 

Examples 

1.    Find  the  square  root  of  .1296  by  logarithms. 

Solution 

Prin.,  log  \/.1296  =  J  log  .1296. 

log.l296  =  9.1126 -10. 

2)19.1126-20 
9.5.563  -  10 

9.5663 -10  =  log. 360. 
.♦.  V.1296  =  .36. 


LOGARITHMS  427 

Find  by  logarithms  the  value  of 

2.  225I  9.    133li  16.    V2.  23.    V2. 

3.  I2.25I  10.    1024^^  17.    V3.  24.    ^:027. 

4.  .2025i  11.    .67242.  I8.    V5.  25.    V30|. 

5.  324^  12.   5.929I  19.    V6.  26.    Vm 

6.  .512I  13.    .46241  20.    ^/2.  27.    Vi52. 

7.  .llSli  14.    1.464li  21.    v4.  28.    v'j032. 

8.  3.375I  15.    .000321  22.    \/3.  29.    V^025. 

Simplify  the  following : 

30.    ^ 35.    IM^^. 

15  X  3.1416  11 


31.  100^  36.    J.434X96* 


4 


48x64x11  >'64xl500 

52=^  x  300  .32  X  5000  x  18 

*    '    12  X  .31225  X  400000'  '   3.14  X  .1222  x  s" 

,,        /       400  38     11x2.63x4.263 
\55x  3.1416  *         48x3.263 

34.    50x?l.  39.       /^ 


/35 


81.63  \  1  06^* 

40.   2^  x  (i)^  X  ^f  X  VI. 

483.    Solution  of  exponential  equations. 

Exponential  equations,  or  equations  that  involve  unknown 
exponents,  are  solved  by  the  aid  of  the  principle  that,  in  any 
system,  equal  numbers  have  equal  logarithms. 

In  simple  cases  the  solution  of  such  equations  may  be  per- 
formed by  inspection,  but  in  general  it  is  necessary  to  use  a  table 
of  logarithms. 

Examples 

1.   Find  the  value  of  x  in  the  equation  2""  =  32  V2, 


428  LOGARITHMS 

Solution 

2«  =  32V2  =  2622  =  2^; 
therefore,  log  (2^)  =  log  (2^), 

or  a;  log  2  =  ^^  log  2, 

.-.  X  =  V- 

2.    Find  the  value  of  x  in  the  equation  2"=  =  48. 

Solution 
Taking  the  logarithm  of  each  member, 

X  log  2  =  log  48. 


log   2 
1.6812 


=  5.58+. 


0.3010 

3.  Find  the  value  of  x  in  the  equation  3^^  —  20  •  3^  +  99  =  0. 

Solution 
Factoring  the  given  equation, 

(3'^-9)(3»'-ll)=0. 

.-.  3^  =  9  or  11. 

Solving  the  equation  3»=  =  9  by  inspection,  since  9  =  3*, 

x-2. 

Taking  the  logarithm  of  each  member  of  3*  =  11, 

xlog3  =  log  11. 

.^^logli^  1,0414  ^2.18+. 
log   3      0.4771 

Therefore,  the  value  of  x  is  either  2  or  2. 18+. 

4.  Given  x^  =  y^  and  x"  =  y^,  to  find  x  and  y. 

Solution 

Raising  the  members  of  the  first  equation  to  the  arth  power,  and  those  of 
the  second  equation  to  the  3d  power, 

and  x^y  =  y^'. 

Hence,  by  inspection,  2x  —  Sy. 

Squaring,  since  4  x^  =  4  y^,    4  x^  =  9  y^  =  4  y», 
.:  y  =  0  or  |, 
and  X  =  0  or  ^. 


LOGARITHMS 


429 


5.    Given  3^  =  27/  and  2"  =  y,  to  find  x  and  y. 

Solution 

3^  =  2  y. 

2^  =  ?/. 

(1.5)^  =  2. 

/.x  log  1.5  =  log  2. 

^   log  2   ^0.3010 
*~  log  1.5     0.1761* 

log  X  ^  0.2328  ; 

X  =  1.709. 

log  y  —  X  log  2. 

/.  log  log  j^  =  log  X  +  log  log  2 

=  0.2328  +  1.4786 

=  1.7114. 

logy  =  0.5145; 

y  =  3.270. 


Dividing  (1)  by  (2), 

Hence,  by  tables, 

By  logarithms, 
♦whence,  by  tables, 
From  (2), 

by  (4)  and  tables. 

Hence,  by  tables, 
whence, 


(1) 
(2) 


(3) 

(4) 
(5) 


Solve  the  following ; 


6. 
7. 
8. 
9. 
10. 


3'  =  81. 
4-  =  10. 
2^  =  80., 

2'"=  512, 


12.  3-'^  -  36-3^  +  243  =  0. 

13.  log  log  a;  =  log  2. 


14 


11.    (2^  =  256. 


15. 
16. 


I  4^  =  20  y. 
2^=  512. 
5^'=  625. 


17. 


18. 


19. 


r  2^+^  =  6, 
1  2^+1  =  3^. 

4^+"  =32, 
22x-»  ^  4_ 

[x  =l+logy. 


484.  Logarithms  applied  to  the  solution  of  problems  in  compound 
interest  and  annuities. 

1.  What  is  the  amount  of  f  1  for  1  year  at  6%  ?  By  what 
must  the  amount  for  1  year  be  multiplied  to  find  the  amount  for 
2  years  at  6%  compound  interest? 

2.  By  what  must  the  amount  for  2  'years  be  multiplied  to 
obtain  the  amount  for  3  years,  compound  interest?  By  what 
must  the  amount  for  3  years  be  multiplied  to  obtain  the  amount 
for  4  years,  compound  interest? 


430  LOGARITHMS 

3.  Since  the  amount  of  any  principal  at  6%  compound  interest 
for  1  year  is  1.06  times  the  principal ;  for  two  years,  1.06  x  1.06, 
or  1.06^  times  the  principal ;  for  3  years,  1.06  x  1.06  x  1.06,  or 
1.06^  times  the  principal,  etc.,  what  will  be  the  amount  {A)  of  any 
principal  (P)  for  n  years  at  any  rate  per  cent  (r)  ? 

Formula.  A  =  P(l  +  ?•)". 

Expressing  the  formula  by  logarithms, 

log  ^  =  log  P  +  ?i  log(l  +  r).  (1) 

.-.  log  P  =  log  ^  -  71  log(l  +  r) ;  (2) 

also  log(l  +  r)=^°g^-^"g-^. 


and  log^-logP 


(3) 
(4) 


Examples 


1.  What  is  the  amount  of  $475  for  10  years  at  6%  compound 
interest  ? 

Solution 

A  =  P(l  +  r)«. 
log475  =  2.6767 
log'l.Oeio  =  0.2530 
\ogA        =  2.9297 
.-.^  =  $850.60. 

2.  What  will  be  the  amount  of  $  225  loaned  for  5  years  at  8% 
compound  interest  ? 

3.  Find  the  amount  of  $700  loaned  for  5  years  at  6%  com- 
pound interest. 

4.  Find  the  amount  of  $400  for  10  years  at  3%  compound 
interest. 

5.  Find  the  amount  of  $1200  for  20  years  at  4%  compound 
interest. 

6.  What  principal  will  amount  to  $1000  in  10  years  at  5% 
compound  interest? 


LOGARITHMS  431 

7.  What  sum  of  money  invested  at  4  %  compound  interest, 
payable  semiannually,  will  amount  to  $  743  in  10  years  ? 

8.  What  principal  loaned  at  4%  compound  interest  will 
amount  to  $  1500  in  10  years  ? 

9.  What  sum  of  money  invested  at  4  %  compound  interest 
from  a  child's  birth  until  he  is  21  years  old  will  yield  f  1000  ? 

10.  In  what  time  will  $800  amount  to  f  1834.50,  if  put  at 
compound  interest  at  5  ^J)  ? 

11.  What  is  the  rate  per  cent  when  $  300  loaned  at  compound 
interest  for  6  years  amounts  to  $  402? 

12.  A  man  agreed  to  loan  $  1000  at  6  %  compound  interest  for 
a  time  long  enough  for  the  principal  to  double  itself.  How  long 
was  the  money  at  interest  ? 

485.  A  sum  of  money  to  be  paid  periodically  for  a  given  num- 
ber of  years,  during  the  life  of  a  person,  or  forever,  is  called  an 
Annuity. 

The  payments  may  be  made  once  a  year,  or  twice,  or  four  times 
a  year,  etc. 

Interest  is  allowed  upon  deferred  payments. 

486.  To  find  the  amount  of  an  annuity  left  unpaid  for  a  given 
number  of  years,  compound  interest  being  allowed. 

1.  If  an  annuity  of  a  dollars  is  not  paid  at  the  end  of  the  first 
year,  how  much  is  then  due  ? 

2.  Upon  what  sum  will  compound  interest  be  computed  for 
the  second  year  ?  What  will  be  the  amount  of  that  sum,  if  the 
rate  is  r  ?  What  will  be  the  whole  sum  due  at  the  end  of  the 
second  year  ?  Ans.  a  4-  a(l  +  r). 

3.  Upon  what  sum  will  compound  interest  be  computed  for 
the  third  year?  What  will  be  the  amount  of  that  sum  at  the 
given  rate  ?  Ans.  a  (1  +  r)  +  a  (1  +  r)^. 

What  will  be  the  whole  sum  due  at  the  end  of  the  third  year  ? 

Ans.  a  +  a  (1  +  r)  +  a  (1  +  rf. 


432  LOGARITHMS 

4.  What  will  be  the  whole  sum  due  at  the  end  of  the  fourth 
year  ?  What  will  be  the  whole  sum  due  at  the  end  of  the  nth. 
year? 

487.  Let  a  represent  the  annuity,  n  the  number  of  years,  ?•  the 
rate,  and  A  the  whole  amount  due  at  the  end  of  the  nth  year. 

Then, 

A=  a  +  a(l  +  r)  +  a(l  -{- rf  -\-  a(l  +  ry  -\ \-a(l-\- r)"-^ 

=  a\l  +  (l  +  r)  +  (1  +  r)2+  (1  +r)3  +  ...  +  (1  +  r)"-'|. 

Since  the  terms  of  A  form  a  geometrical  progression  in  which 
1  4-  »*  is  the  ratio,  §  370,  the  sum  of  the  series  is 

A  =  ^[{l  +  ry-ll 

488.  Sometimes  annuities,  drawing  interest,  are  not  payable 
until  after  a  certain  number  of  years.  It  is  often  necessary, 
therefore,  to  find  the  present  value  of  such  annuities. 

489.  A  sum  that  will  amount  to  the  value  of  an  annuity,  if 
put  at  interest  at  the  given  rate  for  the  given  time,  is  called  the 
Present  Value  of  the  annuity. 

490.  1.  If  P  denotes  the  present  value  of  an  annuity  due  in  n 
years,  allowing  r  %  compound  interest,  to  what  sum  will  P  be 
equal  in  that  time  at  the  given  rate  ?  Ans.  P(l  +  r)". 

2.  Since  the  amount  of  the  present  value  put  at  interest  for 
the  given  time  at  the  given  rate  is  equal  to  the  amount  of  the 
annuity  for  the  same  time  and  rate,  equate  the  two  sums  and  find 
the  value  of  P. 

P(l  +  r)"  =  ?[(l  +  r)"-l]. 


r         (1  +  r)» 


LOGARITHMS  433 

Examples 

1.  What  will  be  the  amount  of  an  annuity  of  $100  remaining 
unpaid  for  10  years  at  6  %  compound  interest  ? 

Solution.  ^  =  -  [(1  +  r)»  -  1]. 

r 

By  logarithms,  1.06"  =  1.7904 

.-.  1.0610  -  1  =    .7904 
log  100  =  2.0000 

log  .7904  =  9.8978  -  10 

colog  .06  =  1.2218 
.-.  logyl  =  3.1196 

Hence,  -4  =  f  1317,  the  amount  of  the  annuity. 

2.  What  is  the  present  value  of  an  annuity  of  f  100  to  continue 
10  years  at  6  %  compound  interest  ? 

Solution.  p  ^<1  AA+l)!^!. 

r         (1  +  r)" 

By  logarithms,  1 .06^  =  1. 7904 

.-.  1.0610  _  1  ^    .7904 

log  100  =  2.0000 

log  .7904  =  9.8978  -  10 

colog  .06  =  1.2218 

colog  1.0610  ^  9.7470  -  10 
.'.  log  P  =  2.8666 

Hence,  P  =  $  735.50,  the  p.  v.  of  the  annuity. 

3.  To  what  sum  will  an  annuity  of  $25  amount  in  20  years  at 
4  ffo  compound  interest  ? 

4.  What  is  the  present  value  of  an  annuity  of  f  300  for  5  years 
at  4  %  compound  interest  ? 

5.  What  will  be  the  amount  of  an  annuity  of  f  17.76  remain- 
ing unpaid  for  25  years  at  3|^  %  compound  interest  ? 

6.  What  is  the  present  value  of  an  annuity  of  f  1000  to  con- 
tinue 20  years,  allowing  compound  interest  at  4i  %  ? 

7.  What  annuity  will  amount  to  f  1000  in  10  years  at  5% 
compound  interest  ? 

ADV.   ALG.  — 28 


PERMUTATIONS   AND    COMBINATIONS 


491.  All  the  different  orders  in  which  it  is  possible  to  arrange 
a  given  number  of  things,  taking  either  some  or  all  of  them  at 
a  time,  are  called  the  Permutations  of  the  things. 

Thus,  the  permutations  of  the  letters  a  and  h  are  ah,  ha  ;  the  permuta- 
tions of  three  letters,  two  at  a  time,  are  ah,  ac,  ha,  be,  ca,  cb. 

492.  All  the  different  selections  that  it  is  possible  to  make 
from  a  given  number  of  things,  taking  either  some  or  all  of  them 
at  a  time,  without  regard  to  the  order  in  which  they  are  placed, 
are  called  the  Combinations  of  the  things. 

Thus,  while  the  permutations  of  three  letters,  two  at  a  time,  are  ah  and 
ba,  be  and  cb,  and  ca  and  ae,  their  comhinations,  two  at  a  time,  are  ah  (or  -7. 
ha),<  be  (or  cb),  and  ac  (or  ca)  ;  again,  the  six  permutations  of  three  letters 
among  themselves,  viz.,  ahc,  acb,  bca,  bac,  cab,  and  cha,  form  but  one  com- 
bination, abc  (or  acb,  or  bca,  or  bac,  or  cab,  or  cha). 

It  is  evident  that  there  can  be  but  one  combination  of  any  number  of 
things  taken  all  at  a  time. 

493.  Notation.  —  The  symbol  for  the  number  of  permutations 
of  n  different  things,  taken  r  at  a  time,  is  P"^ ;  of  n  different 
things,  taken  w  at  a  time,  or  all  together,  P^. 

.  The  symbol  for  the  number  of  combinations  of  n  different 
things,  taken  r  at  a  time,  is  C" ;  of  n  different  things,  taken  n 
at  a  time,  or  all  together,  is  O^. 

494.  The  product  of  the  successive  natural  numbers  from  1 
to  n,  or  from  n  to  1,  inclusive,  is  called  factorial  n,  written  \n. 

[5  =  1x2x3x4x5,  or  5x4x3x2x1; 

[w  =  1  .  2  .  3  ..•  (n  -  2)(n  -  l)n,  or  w(n  -  l)(n  -  2)(n  -  3)  •••  3  •  2  •  1. 

in  is  sometimes  written  n  ! 

434 


PERMUTATIONS  AND   COMBINATIONS  435 

495.  To  find  the  number  of  permutations  of  n  different  things  taken 
V*  at  a  time. 

Since  the  permutations  of  the  letters  a,  b,  and  c,  taken  2  at  a 
time,  are  ab  and  ac,  ba  and  be,  ca  and  cb,  formed  by  writing  after 
each  of  the  letters  a,  b,  and  c,  each  of  the  other  letters  in  turn,  the 
number  of  permutations  of  3  different  things  taken  2  at  a  time  is 
3x2. 

The  number  of  permutations  of  n  letters  taken  2  at  a  time  may 
be  found  by  associating  with  each  of  the  n  letters  each  of  the 
n  —  1  other  letters.  Consequently,  the  number  of  permutations 
of  n  different  things  taken  2  at  a  time  is  n{n  —  1). 

Since  the  number  of  permutations  of  n  letters  2  at  a  time  is 
n(n  —  1),  if  the  letters  are  taken  3  at  a  time  there  will  be  n  —  2 
letters  each  of  which  may  be  associated  with  each  of  the  n{n  —  1) 
permutations  of  letters  taken  2  at  a  time.  Hence,  the  number  of 
permutations  of  n  different  things  taken  3  at  a  time  is 

n  (n  -  1)  (m—  2). 

Principle  1.  —  The  number  of  permutations  ofn  different  things 
taken  r  at  a  time  is  equal  to  the  continued  product  of  the  natural 
numbers  from  n  to  n  —  (r  —  1)  inclusive.  The  number  of  factors 
is  r.     That  is, 

P"  =  ?i(n  —  1) (w  —  2)  •••  to  r  factors 

=  w(w-l)(n-2)  ...(ji-r  +  l).  (I) 

Multiplying  and  dividing  the  second  member  of  (I)  by 

(w  —  r) (w  —  r  -^  i)  (n  —  r  —  2)  •••  2  •  1 ;  that  is,  by  \n  —  r, 

\n 
P?  =  -l--  (II) 

I  n  —  r 

It  will  usually  be  more  convenient  to  employ  formula  (I)  in 
solving  numerical  examples;  but  when  simply  algebraic  results 
are  desired,  formula  (II)  will  be  preferable. 

496.  When  r  =  n,  that  is,  when  the  things  are  taken  all 
together,  the  last,  or  nth,  factor  in  (I)  is  1.     Consequently, 

Principle  2. —  The  number  of  permutations  ofn  different  things 
taken  all  at  a  time  is  equal  to  \n.     That  is, 

Pj  =  w (w  —  1) (71  —  2)  •••  to  n  factors  =  | w.  (HI) 


436  PERMUTATIONS  AND   COMBINATIONS 

Examples 

1.  Three  boys  enter  a  car  in  which  there  are  5  empty  seats. 
In  how  many  ways  may  they  choose  seats  ? 

Solution.  —  Since  the  first  boy  may  choose  any  one  of  5  seats  ;  and  since, 
after  he  has  chosen  one  of  them,  for  each  seat  that  he  may  choose,  the  second 
boy  may  choose  any  one  of  the  4  seats  remaining,  the  greatest  possible  num- 
ber of  ways  in  which  two  of  the  boys  may  be  seated  is  5  x  4. 

Again,  since  after  each  clioice  of  seats  made  by  two  of  the  boys  there  will 
be  left  to  the  third  boy  a  choice  of  one  of  the  3  seats  remaining,  the  number 
of  ways  in  which  all  may  choose  seats  is  5  x  4  x  3,  or  60. 

Or,  by  (I),  P;?  =  n{n  -  l)(n  -  2)  ...  (?i  -  r  +  1), 

P|  =  5  X  4  X  3  =  GO. 

2.  How  many  numbers  between  100  and  1000  can  be  expressed 
by  the  figures  1,  3,  5  ? 

Solution.  —  Since  the  numbei's  lie  between  100  and  1000,  each  must  be 
expressed  by  three  figures.  Hence,  the  number  of  numbers  between  100  and 
1000  that  can  be  expressed  by  the  figures  1,  3,  and  5  is  the  same  as  the  num- 
ber of  permutations  'of  these  3  figures  taken  3  at  a  time. 

Since,  Prin.  2,         .         P|  =  (3  =  3.2.1=:6, 

there  are  six  such  numbers.     They  are  135,  153,  351,  315,  513,  and  531. 

3.  How  many  permutations  can  be  made  of  the  letters  in  the 
word  Albany,  each  beginning  with  capital  A  ? 

Solution.  —  Since  A  is  to  be  prefixed  to  each  permutation  of  the  5  other 
letters,  the  required  number  is 

P|  =  5x4x3x2xl  =  120. 

4.  In  how  many  orders  may  4  persons  sit  on  a  bench  ? 

5.  How  many  permutations  may  be  made  of  the  letters  in  the 
word  steam  f 

6.  If  10  athletes  run  a  race,  in  how  many  ways  may  the  first 
and  second  prizes  be  awarded  ? 

7.  In  how  many  different  orders  may  the  colors  violet,  indigo, 
blue,  green,  yellow,  orange,  and  red  be  arranged  ? 

%t  There  are  five  routes  to  the  top  of  a  mountain.  In  how 
many  ways  may  a  person  go  up  and  return  by  a  different  way  ? 


^ 


PERMUTATIONS  AND   COMBINATIONS  437 

497.  To  find  the  number  of  combinations  of  n  different  things 
taken  /•  at  a  time. 

Since  two  letters,  as  a  and  h,  have  two  permutations,  ab  and  ba, 
but  form  only  one  combination,  the  number  of  combinations  of  n 
letters  taken  2  at  a  time  is  one  half  the  number  of  permutations 
of  n  letters  taken  2  at  a  time. 

Since  three  letters  taken  3  at  a  time  have  3x2  permutations, 
but  form  only  one  combination,  the  number  of  combinations  of  n 
letters  taken  3  at  a  time  is  obtained  by  dividing  the  number  of 
permutations  of  n  letters  taken  3  at  a  time  by  3  x  2. 

Since  four  letters  taken  4  at  a  time  have  |4  permutations  but 
form  only  one  combination,  to  obtain  the  number  of  combinations 
of  n  letters  taken  4  at  a  time,  the  number  of  permutations  of  n 
letters  taken  4  at  a  time  must  be  divided  by  [4. 

Hence  it  follows  that : 

Principle  3.  —  The  numbei'  of  combinations  of  n  different  things 
taken  r  at  a  time  is  equal  to  the  number  of  permutations  of  n  dif- 
ferent things  taken  r  at  a  time,  divided  by  the  number  of  permuta- 
tions of  r  different  things  taken  all  togetlier.     That  is, 

ri^  —  p"  _t.  p*-  _  '>^{n  —  1)  (n  —  2)  •  •  •  to  r  factors 
?*  (/  —  1)  (r  —  2)  •••  to  r  factors 

^  n  (n  -  1)  (n  -  2)  »•  (w  -  r  +  1)   (IV) 
1  .2  .3  ..•  r 

or,  by  (II), 

(V) 

498.  Since  for  every  combination  of  r  things  out  of  n  different 
things  there  is  left  a  combination  of  n  —  r  things,  it  follows  that : 

Prixciple  4. —  Tlie  number  of  combinations  of  n  different  things 
is  the  same  when  taken  n  —  r  at  a  time  as  when  taken  r  at  a  time. 
That  is, 

(fn-r^Crr^.       ;-      ••  (VI) 

.,     [r  \n  —  r 


1 

n 

\n 

—  r 

\n 

\r 

\n  —  r 

438  PERMUTATIONS  AND   COMBINATIONS 

The  above  principle  may  be  established  as  follows : 
By(V).  c:  =  --}^—  (1) 


Substituting  n  —  r  for  r,     cZ-r  = 


Since  the  second  members  of  (1)  and  (2)  are  identical,  cZ-r  =  Or 


\r\n  —  r 

\n  —  r\n  —{n- 

-r) 

\n  —  r\r 

(2) 


The  above  principle  is  useful  in  abridging  numerical  computations. 
Thus,  the  number  of  combinations  of  18  things  taken  16  at  a  time  is  com- 
puted by  Prin.  3  as  follows  : 

^18  _  18  •  17  .  16  •  15  •  14  •  13  •  12  •  11  •  10  •  9-  8-  7  •  6  •  5  •  4  ■  3  ^  jgg 
^®        1  .  2  .  3  •  4  .  5  •  6  .  7  .  8  .  9  .  10  .  11  .  12  .  13  •  14  .  15  •  16 

By  Prin.  4,  the  computation  is  abridged  as  follows : 

1-2 

Examples 

1.  A  man  has  6  friends  and  wishes  to  invite  4  of  them  to 
dinner.     In  how  many  ways  may  he  select  his  guests? 

Solution.  —  Since  each  party,  or  combination,  of  4  guests  could  be 
arranged,  or  permuted,  in  [4  ways,  the  number  of  combinations  must  be  — 
of  the  number  of  permutations  of  6  things  taken  4  at  a  time.  I— 

Hence,  the  number  of  ways  is 

1x2x3x4  'o^ 

2.  A  man  and  his  wife  wish  to  invite  11  of  their  friends,  6  men   -^ 
and  5  women,  to  dinner,  but  find  that  they  can  entertain  only  8 
guests.     In  how  many  ways  may  they  invite  4  men  and  4  women  ? 

SoLDTioN.  —  As  in  the  previous  example,  4  men  may  be  selected  from  6 
men  in  15  ways,  and  in  a  similar  manner  4  women  may  be  selected  from  6 
women  in  5  ways.  ^ 

Since,  when  any  set  of  4  men  has  been  invited,  the  party  of  8  may  be  com- 
pleted by  inviting  any  one  of  5  sets  of  4  women,  the  whole  number  of  differ- 
ent parties  that  it  is  possible  to  invite  is  15  x  5,  or  75.    That  is, 

r.6,,r.^_6-5.4.3      5.4.3-2_ 


PERMUTATIONS  AND    COMBINATIONS  439 

3.  In  how  many  ways  may  a  baseball  nine  be  selected  from 
12  candidates? 

4.  Find  the  value  of  C'^;  of  (7>|;  of  C^. 

5.  How  many  different  combinations  of  5  cards  can  be  formed 
from  52  cards  ? 

6.  Which  is  the  greater,  C'^  or  C'^  ?     C't  or  G^l  ? 

7.  From  11  Republicans  and  10  Democrats  how  many  different 
committees  can  be  selected  composed  of  6  Republicans  and  5 
Democrats  ? 

8.  A  man  forgets  the  combination  of  figures  and  letters  by 
which  his  safe  is  opened.  If  they  are  arranged  on  the  circum- 
ferences of  three  wheels,  one  bearing  the  numbers  0  to  9  inclu- 
sive, another  the  letters  A  to  M  inclusive,  and  the  third  the  letters 
N  to  2  inclusive,  what  is  the  greatest  number  of  trials  he  may 
have  to  make  to  open  the  safe  ? 

9.  From  6  consonants  and  4  vowels  how  many  words  may  be 
formed  each  consisting  of  4  consonants  and  2  vowels,  if  any 
arrangement  of  the  letters  is  considered  a  word  ? 

Solution.  — The  number  of  combinations  is  Ct  x  C\;  and  since  by  per- 
muting the  letters  of  each  combination  |^  words  can  be  formed,  the  number 
of  words  is  C\x  G\x  [6. 

10.  In  an  omnibus  that  will  seat  8  persons  on  a  side  there  are 
seated  4  persons,  3  on  one  side  and  1  on  the  other.  In  how  many 
ways  may  12  more  persons  be  seated  ? 

Solution.  —  Since  5  persons  must  take  seats  on  one  side  and  7  persons  on 
the  other,  12  persons  are  to  be  divided  into  two  classes,  6  and  7.  The  number 
of  these  combinations,  formula  (V),  is 

C'/,orC?,=|i. 

Since  each  combination  of  5  may  have  [5  permutations  of  the  5  that 
compose  it,  and  each  combination  of  7  may  have  [7  permutations  each  of 
which  may  be  associated  with  each  of  the  [5  permutations,  the  required 
number  of  ways  is  C^i  >j  P5  x  P7, 

(12 
or  J=x|5x  7=112. 

Or,  since  there  are  12  persons  to  be  seated  in  12  seats,  the  number  of  ways 


i8Pii  =  |12. 

\*  A 

^1 ;  •  'A  .> 

440  PERMUTATIONS  AND   COMBINATIONS 

11.  Out  of  20  consonants  and  5  vowels  how  many  words  con- 
taining 3  consonants  and  3  vowels  can  be  formed,  if  any  arrange- 
ment of  the  letters  is  considered  a  word  ? 

12.  How  many  different  sums  can  be  paid  with  a  cent,  a  half- 
dime,  a  dime,  a  quarter,  and  a  dollar  ? 

13.  From  5  boys  and  5  girls  how  many  committees  of  6  can  be 
selected  so  as  to  contain  at  least  2  boys  ? 

14.  A  company  of  a  soldiers  is  joined  by  another  company  of 
h  soldiers.  In  how  many  ways  is  it  possible  to  leave  c  of  them  to 
garrison  the  fort,  dividing  the  rest  into  two  scouting  parties,  one 
of  m,  the  other  of  n  soldiers  ? 

15.  If  Cs  =  2  Cl,  find  the  number  of  things. 

Solution.—  By  formula  (V),   CI  =  ,^,^       and  CI  =  — -t^^ 

|2|w  —  2 


Since  01  =  2,  CI 


2> 


\n-2 


(V), 

CI: 

|5|u- 

6 

\n 

2[n 

\2\n- 
1 

[5\n- 
1 

-_5 

2 

|6|n- 

-6 

\n-2 
=  \5\n- 

.:{n^ 

-2: 

5. 

or  (n-2)(n-3)(n-4)=5  x4x3x2xl, 


|w-5 

or  (n-2)(n-3)(tt-4)  =6  X  5  X  4. 

.-.  n  =  8. 
16.   If  3  C^  =  2  C"t\  find  n,  Cl  and  (7"t\ 

499.   To  find  the  number  of  circular  permutations  of  n  different 
things  taken  n  at  a  time. 

Suppose  four  letters  a,  b,  c,  d,  placed  in  a  fixed  position  around 
a  circle  in  the  order  abed.  Since  the  arrange- 
ment may  be  read  abed,  bcda,  cdab,  or  dabc, 
without  changing  the  direction  in  which  the 
letters  are  read,  it  is  evident  that  each  circu- 
lar permutation  of  4  letters  taken  all  together 
takes  the  place  of  4  permutations  of  the 
letters  all  together. 

That  is,  the  number  of  circular  permuta- 
tions of  4  things  taken  all  together  is  one  fourth  of  the  number  of 
permutations  of  4  things  taken  all  together. 


PERMUTATIONS  AND   COMBINATIONS  44l 

The  whole  number  of  permutations  of  n  things  taken  all 
together  is  [n.  But  if  the  n  things  are  arranged  around  a  circle, 
n  of  these  permutations  may  be  obtained  from  any  circular  per- 
mutation without  disturbing  the  relative  positions  of  the  things. 

Hence, 

Principle  5.  —  The  number  of  circular  permutations  of  n  things 

taken  all  together  is  equal  to  -th  of  the  whole  number  of  their  per- 

n 

mutations  taken  all  together.     That  is, 

In 
i^* (circular)  =  =  =  \n -1.  (VII) 

n      


Examples 

1.  In  how  many  orders  may  6  persons  seat  themselves  around 
a  table  ? 

2.  In  how  many  orders  may  4  gentlemen  and  their  wives  seat 
themselves  around  a  table  ? 

3.  In  how  many  orders  may  4  gentlemen  and  their  wives  seat 
themselves  around  a  table  so  that  each  gentleman  sits  opposite 
his  wife  ? 

4.  In  how  many  orders  may  4  gentlemen  and  their  wives  seat 
themselves  around  a  table  so  that  each  gentleman  sits  opposite 
a  lady  ? 

5.  In  how  many  ways  may  the  colors  violet,  indigo,  blue,  green, 
yellow,  orange,  and  red  be  arranged  on  a  disk,  the  colors  radiating 
from  the  center  ? 

500.  To  find  the  number  of  permutations  of  n  things  taken  n  at 
a  time  when  they  are  not  all  different. 

If,  in  the  permutation  (a,  b,  c,  d,  e,  f  g),  the  letters  h,  d,  and  g 
are  permuted  while  the  other  letters  remain  fixed  in  position,  the 
resulting  number  of  permutations  will  be  the  same  as  the  number 
of  permutations  of  b,  d,  and  g.  If  b,  d,  and  g  are  different  things, 
the  number  of  permutations  resulting  will  be  [3 ;  but  if  6,  d,  and  g 
become  alike,  there  will  be  but  1  permutation. 

That  is,  the  number  of  permutations  of  any  number  of  things 


442  PERMUTATIONS  AND   COMBINATIONS 

when  three  of  them  are  alike  is  equal  to  the  number  of  permuta- 
tions of  the  things,  considered  as  all  different,  divided  by  |3  ;  if  4 
of  the  things  are  alike,  by  [4 ;  ii  p  oi  the  things  are  alike,  by  [p. 
Hence,  it  follows  that : 

Principle  6.  —  The  number  of  permutations  of  n  things,  taken 

\n 
all  together,  when  p  of  them  are  alike,  is  —  • 

[P 
Since,  if  q  of  the  remaining  n—p  different  things  become  alike, 
but  different  from  the  p  like  things,  the  number  of  permutations 
must  be  divided  by  [g ;  if  r  others  become  alike,  by  [r ;  etc. :  it 
follows  that : 

Principle  7.  —  Tfie  number  of  permutations  of  n  things,  taken 
all  together,  when  p  of  them  are  of  one  kind,  q  of  another,  r  of 

another,  etc.,  is  ■ — -= 

\p\q\r-- 

Examples 

1.  How  many  permutations  may  be  made  with  the  letters  of 
the  word  Mississippi  taken  all  together  ? 

Solution.  — The  number  is     —    =  34650. 

[4  [4  [2 

2.  How  many  permutations  may  be  made  with  the  letters  of 
^ch  of  the  following  words,  all  at  a  time  in  each  case :  zoology, 
coefficient,  ecclesiastical,  divisibility  ? 

3.  How  many  permutations  may  be  made  with  the  letters  rep- 
resented in  the  product  a*b^c^  written  out  in  full  ? 


501.    To  find  the  total  number  of  combinational  n  different  things. 

The  number  of  combinations  of  n  different  things  taken  succes- 
sively 1,  2,  3,  •••  w  at  a  time  is  called  the  total  number  of  combi- 
nations of  n  things. 

The  total  number  of  combinations  of  2  things  is 

Cf+ 01=2  +  1  =  3,  or  22-1. 
The  total  number  of  combinations  of  3  things  is 

Of  +  Ci  +  C'i  =  3  +  3  -H  1  =  7,  or  23  -  1. 


PERMUTATIONS  AND   COMBINATIONS  443 

The  total  mimber  of  combinations  of  4  things  is 

Ct  +  q  +  C|  +  C^4  =  4  +  6  +  4  +  1  =  15,  or  2*  -  1. 

Principle  8.  —  The  total  number  of  combinations  of  n  different 
'things  is  2"  —  1. 

The  above  principle  may  be  established  as  follows: 

§  452,  when  n  is  a  positive  integer, 

(1  +  x)»  =  l  +  nx  +  »(»-!) a;2  +  ...  +  n(n  -  l)(n  -  2)  ...  1^„ 
1-2  1  ■  2  •  3  .••  n 

Ifx  =  l,  2n  =  1  +  n  +  "("  -  ^)  +  .-  +  '<""  -  ^)('^  -  ^)  -  ^ 

1.2  1  .2.3...ri 

Prin.  3,  =i+Gl  +  C';+  ...  +  C',:  =  1  +  C;.^. 


'^  total  -  "^ 


Examples 


1.  How  many  different  sums  can  be  paid  with  a  cent,  a  3-cent 
piece,  a  half-dime,  a  dime,  a  quarter,  and  a  dollar  ? 

Solution.  —  Total  0^  =  2^-1  =  63. 

2.  A  man  has  10  friends.  In  how  many  ways  may  he  invite 
one  or  more  of  them  to  dinner  ? 

3.  How  many  different  quantities  can  be  weighed  by  weights 
of  1  oz.,  1  lb.,  i  lb.,  5  lb.,  and  10  lb.  ?  ^ 

4.  How  many  signals  can  be  made  with  7  flags  ? 

5.  By  permuting  the  letters  of  the  word  counter,  how  many 
permutatiatis  ^i  be  formed 

(a)  ending  in  er  ? 

(b)  with  n  as  the  middle  letter  ? 

(c)  without  changing  the  position  of  any  vowel  ? 

(d)  beginning  with  a  consonant  ? 

6.  How  many  numbers  can  be  formed  with  the  digits  1,  2,  3, 
4,  3,  2,  1,  so  that  the  odd  digits  always  occupy  the  odd  places? 

7.  If  the  number  of  permutations  of  n  different  things  taken 
5  at  a  time  is  equal  to  24  times  the  number  of  permutations  of 
the  same  number  of  things  taken  2  at  a  time,  find  n. 


444  PERMUTATIONS   AND   COMBINATIONS 

8.  How  many  permutations  can  be  made  of  the  letters  of  the 
word  international,  taken  all  together  ? 

9.  A  man  has  five  coats,  six  vests,  and  eight  pairs  of  trousers. 
In  how  many  different  suits  may  he  appear  ? 

10.  In  how  many  ways  may  a  committee  of  5  men  and  3 
women  be  chosen  from  10  men  and  5  women  ?  | 

11.  Mr.  Alexander  has  5  pairs  of  gloves.  In  how  many  ways 
can  he  select  a  right-hand  glove  and  a  left-hand  glove  without 
selecting  a  pair  ? 

12.  In  how  many  ways  can  a  party  of  12  young  people  form  a 
ring? 

13.  In  how  many  ways  can  5  ladies  and  5  gentlemen  be  seated 
at  a  round  table  so  that  no  two  ladies  sit  together  ? 

14.  A  man  belongs  to  a  club  of  16  members  and  every  day  he 
invites  three  members  to  dine  with  him.  For  how  many  days 
can  he  invite  a  different  party  each  day  ? 

15.  In  how  many  ways  may  12  persons  seat  themselves  at  two 
round  tables,  6  persons  at  each  table  ? 

16.  In  how  many  ways  may  8  books  be  arranged  on  a  shelf  so 
that  two  particular  books  will  not  be  together  ? 

17.  In  how  many  ways  may  a  committee  of  5  be  appointed 
from  12  boys  and  10  girls  so  that  there  will  be  more  boys  than 
girls  in  the  committee  ? 

18.  A  boat's  crew  consists  of  8  men,  of  whom  3  can  row  only 
on  the  port  side  and  2  only  on  the  starboard  side.  In  how  many 
ways  can  the  crew  be  seated  ? 

19.  In  how  many  different  numbers  less  than  1000  does  the 
digit  9  appear  ? 

20.  From  a  detachment  of  20  soldiers  4  are  detailed  each  night 
for  picket  duty.  For  how  many  successive  nights  can  a  different 
picket  be  formed  and  how  many  times  will  each  soldier  be  on 
duty  during  this  interval  ? 


PROBABILITY 


502.  Suppose  that  a  person  draws  one  ball  at  random  from  a 
bag  containing  3  white  balls  and  2  black  balls,  and  it  is  required 
to  find  his  chance  of  drawing  a  white  ball. 

Since  this  event  can  happen  in  3  ways,  namely,  by  drawing 
any  one  of  the  3  Avhite  balls,  and  can  fail  in  2  ways,  namely,  by 
drawing  either  of  the  2  blacft  balls,  it  is  seen  that  the  ratio  of 
the  number  of  ways  favorable  to  drawing  a  white  ball  to  the 
whole  number  of  ways,  favorable  and  unfavorable,  is  3:5.  This 
is  expressed  by  saying  that  the  chance,  or  prohability,  of  drawing 
a  white  ball  is  f . 

Similarly,  the  probability  of  drawing  a  black  ball  is  |. 

If  one  of  the  black  balls  is  removed  before  drawing,  the  proba- 
bility of  drawing  a  white  ball  is  increased  to  f  and  the  probability 
of  drawing  a  black  ball  is  diminished  to  \.  If  both  black  balls 
are  removed,  the  probability  of  drawing  a  white  ball  amounts  to 
certainty,  expressed  by  |,  or  1,  and  the  probability  of  drawing  a 
black  ball  becomes  zero,  the  occurrence  being  impossible. 

503.  If  an  event  can  happen  in  a  ways  and  fail  in  b  ways,  and 
all  of  these  ways  are  equally  likely  to  occur,  the  chance,  or  proba- 
bility, that  it  will  happen  is  r,  and   the  probability  that  it 

1  Ct  ~p"  0 

will  fail  is  r^- 

a  -\-  b 

It  a  =  b,  the  chances  are  said  to  be  even. 

504.  The  ratio  of  the  probability  that  an  event  will  happen  to 
the  probability  that  it  will  fail  is  called  the  odds  in  favor  of  the 
event,  and  the  reciprocal  ratio  is  called  the  odds  against  the 
event. 

445 


446  PROBABILITY  ^ 


\ 


Thus,  if  a  person  draws  one  ball  at  random  from  a  bag  containing  3  white 
balls  and  2  black  balls,  the  odds  are  3  to  2  in  favor  of  his  drawing  a  white 
ball  and  2  to  3  against  his  drawing  a  white  ball. 

505.  Principles.  — 1.  If  the  probability  that  an  event  will  hap- 
pen is  equal  to  p,  the  probability  that  it  will  fail  is  equal  to  1  —  p. 

2.  If  an  event  is  certain  to  happen,  its  probability  is  1 ;  if  an  event 
is  certain  to  fail,  its  probability  is  zero. 

The  above  principles  may  be  established  as  follows : 
Suppose  that  an  event  can  happen  in  a  ways  and  fail  in  h  ways. 

1.  Then,  by  the  definition  of  probability,  the  probability  that  the  event 

will  happen  is and  the  probability  that  it  will  fail  is .     Since 

the  sum  of  these  probabilities  is  1,  if  the  probability  that  the  event  will  hap- 
pen is  equal  to  p,  the  probability  that  it  will  fail  is  equal  to  1  —  p. 

2.  If  an  event  is  certain  to  happen,  the  probability  that  it  will  fail  is 
equal  to  zero  ;  that  is,  1  —  p  =  0,  or  p  =  1 . 

If  the  event  is  certain  to  fail,  the  probability  that  it  will  fail,  as  has  just 
been  proved,  is  equal  to  1 ;  that  is,  1  —  p  =  1,  orp  =  0. 


Problems 

1.  If  one  ball  is  drawn  at  random  from  a  bag  containing  3 
white  balls,  4  black  balls,  and  5  red  balls,  what  is  the  probability 
that  it  is  red  *?     What  are  the  odds  against  its  being  red  ? 

Solution 

Since  5  of  the  12  balls  are  red,  the  probability  of  drawing  a  red  ball  is 
5  in  12,  or  ^j. 

Since  7  of  the  12  balls  are  not  red,  the  probability  that  the  ball  drawn  is 
not  red  is  ^^,  and  the  odds  against  drawing  a  red  ball  are  ^^  to  ^,  or  7  to  5. 

2.  If  three  coins  are  tossed,  what  is  the  probability  (a)  that  all 

fall  heads  ?     (6)  that  a  certain  two  fall  heads  and  the  third  tail  ? 

(c)  that  any  two  fall  heads  and  one  tail  ?     (d)  that  at  least  two 

fall  heads  ? 

Solution 

For  convenience  call  the  coins  A,  B,  and  C. 


PROBABILITY  -     447 

(a)  Since  the  probability  that  any  coin  falls  head  is  J,  and  since,  whether 
A  falls  head  or  tail,  B  may  fall  either  head  or  tail,  the  probability  that  both 
A  and  B  fall  heads  is  ^  x  ^,  or  \.  Again,  since  when  both  A  and  B  fall 
heads  C  may  fall  either  head  or  tail,  the  probability  that  all  fall  heads  is 

(b)  Since  the  probability  that  a  certain  two,  as  A  and  B,  fall  heads  is 
^  X  ^,  or  \,  and  since  when  A  and  B  have  fallen  heads  C  may  fall  either  head 
or  tail,  the  probability  that  a  certain  two  fall  heads  and  the  third  tail  is  |  x  |, 
orf 

(c)  Since  A  and  B,  or  A  and  C,  or  B  and  C  may  fall  heads  while  in  each 
case  the  third  coin  has  even  chances  of  falling  head  or  tail,  the  probability 
that  some  two  coins  will  fall  head  and  the  third  coin  tail  is  3  times  as  great 
as  the  probability  that  a  certain  two  will  fall  heads  and  the  third  tail. 
Hence,  by  (fc)  the  probability  of  this  event  Ls  |. 

(d)  The  probability  that  at  least  two  coins  fall  heads  is  greater  than  the 
probability  that  two  fall  heads  and  one  tail,  for  the  former  includes,  while 
the  latter  excludes,  the  probability  that  all  fall  heads.  Since  by  (c)  the 
probability  that  two  fall  heads  and  one  tail  is  |,  and  by  (a)  the  probability 
that  all  fall  heads  is  |,  the  probability  that  at  least  two  coins  fall  heads  is 
I  +  i,  or  ^• 

3.  From  a  bag  containing  8  white  balls  and  4  black  balls,  4 
balls  are  drawn  at  random.  What  is  the  probability  that  2  are 
white  and  2  are  black  ? 

First  Solotion.  —  Suppose  that  the  balls  are  drawn  one  at  a  time.  The 
probability  that  the  first  drawn  is  white  is  /^ ;  if  the  first  is  white,  the  proba- 
bility that  the  second  is  white  is  j'^j^ ;  when  two  white  balls  have  been  drawn, 
the  probability  that  the  third  is  black  is  ^ ;  and  when  2  white  balls  and  1 
black  ball  have  been  drawn,  the  probability  that  the  next  is  black  is  |. 

Hence,  the  probability  of  drawing  2  white  balls  and  2  black  balls  in  the 

14 
order  WWBB  is  /^  •  t't  •  i^u  •  I )  ^^^  similarly  for  each  of  the  -=-  orders  in 
which  2  white  balls  and  2  black  balls  can  be  drawn.  I—  L- 

(8-7-4-3)C4-3.2 -1)     ^56 
'■  ^"(12.11  •10.9)(1.2.  1-2)      165' 

Second  Solutiov.  — From  12  balls  4  balls  can  be  selected  in  C^  ways. 
Again,  from  8  white  balls  2  white  balls  can  be  selected  in  G\  ways,  and  from 
4  black  balls  2  black  balls  can  be  selected  in  C\  ways. 

Therefore,  out  of  C^  ways  of  selecting  4  balls  from  12,  C\  •  C\  lead  to" 
selecting  2  white  balls  and  2  black  balls. 

.   D  =  ^il_^2=    (R-7)(4.3)(1.2.3.4)    ^  ^6 
'  C^         (1.2)(1.2)(12.11.10.9)      165 


448  PROBABILITY 


J 


4.  A  and  B  and  eight  other  persons  seat  themselves  at  random 
around  a  circular  table.  What  is  the  probability  that  A  and  B 
sit  together  ? 

First  Solution 

Since  A  may  sit  with  each  of  the  9  other  persons  in  2  ways,  on  the  right 
or  on  the  left,  but  whenever  he  sits  on  the  right  of  one  person  he  sits  on  the 
left  of  another,  the  whole  number  of  ways  in  which  he  can  sit  with  different 
persons  is  9.  Now  on  2  occasions  A  sits  with  B,  once  on  B's  right  and  once 
on  B's  left.  Hence,  if  the  ten  persons  seat  themselves  at  random,  the  prob- 
ability that  A  and  B  will  sit  together  is  \. 

Second  Solution 

If  A  and  B  and  the  other  8  persons  seat  themselves  at  random,  §499,  the 
whole  number  of  ways  in  which  they  may  sit  around  the  circular  table  is  [9. 

To  find  how  many  of  these  ways  lead  to  seating  A  and  B  together,  sup- 
pose that  A  and  B  first  seat  themselves  side  by  side  while  the  remaining 
persons  take  at  random  the  8  remaining  seats.  A  and  B  may  sit  together  in 
two  ways,  A  on  the  right  or  on  the  left  of  B,  and  with  each  of  these  ways  the 
other  persons  may  be  seated  in  any  one  of  F%  ways.  Hence,  2  T%  ways  lead 
to  seating  A  and  B  together. 

Since  there  are  2  Pg,  or  2  [8,  ways  that  lead  to  seating  A  and  B  together 
and  |9  ways  in  all,  the  probability  that  A  and  B  sit  together  is 

2J|_2J8_2 
|9~9|8~9" 

5.  In  a  hat  are  placed  25  tickets  numbered  from  1  to  25. 
Three  tickets,  each  entitling  the  holder  of  their  duplicates  to  a 
prize,  are  drawn  from  the  hat.  What  is  the  probability  that  A, 
wno  holds  a  duplicate  of  number  10  and  also  of  number  li,  will 
receive  at  least  one  prize  ? 

Solution 
The  whole  number  of  ways  in  which  3  tickets  can  be  drawn  is 


-,25  _  25  •  24  •  2.3 
'^       1.2.3 


=  2300. 


To  find  how  many  of  the  2300  possible  ways  of  drawing  3  tickets  involve 
the  drawing  of  any  particular  ticket,  as  ticket  number  10,  suppose  that  all 
the  groups  of  3  tickets  in  which  ticket  number  10  is  found  are  formed  by  first 
selecting  this  ticket  and  then  associating  with  it  every  possible  combination 
of  the  remaining  24  tickets  taken  2  at  a  time.  From  this  it  is  evident  that 
any  particular  ticket  is  found  in  C^  of  the  2300  possible  groups  of  3  tickets. 


PROBABILITY  449 

Hence,  the  probability  that  A  will  draw  a  prize  with  the  duplicate  of  num- 
ber 10  is 

O^  ^  12  •  23  ^  3 
2300       2300       25' 

Since  he  has  the  same  chance  with  the  duplicate  of  number  11,  his  chance 
of  drawing  one  prize  is  f^. 

In  like  manner,  since  any  two  particular  tickets,  as  numbers  10  and  11,  are 
found  in  Cf  of  the  2300  possible  groups  of  3  tickets,  the  probability  that  A 
will  draw  two  prizes  is 

Cf  ^    1 
2300      100* 

Since  the  probability  that  A  will  draw  at  least  one  prize  includes  not  only 
the  probability  that  he  will  draw  a  prize  with  one  or  the  other  of  the  two 
duplicates,  but  also  the  probability  that  he  will  draw  prizes  with  both  of  them, 
the  probability  that  he  will  draw  at  least  one  prize  is  /^  +  ^^g,  or  \. 

6.  From  a  bag  containing  4  white  balls  and  8  red  balls  one  ball 
is  taken  at  random.  What  is  the  probability  that  it  will  be  white  ? 
What  are  the  odds  against  drawing  a  white  ball  ? 

7.  A  die  has  2  white  sides  and  4  black  sides.  What  is  the 
probability  that  a  white  side  will  be  thrown  ? 

8.  When  two  dice  are  thrown,  what  is  the  chance  of  throwing 
double  sixes  ?  of  throwing  a  5  and  a  6  ? 

9.  If  two  cards  are  drawn  at  random  from  a  pack  of  52  cards, 
4  of  which  are  kings,  what  is  the  probability  that  both  are  kings  ? 

10.  A  committee  of  four  is  chosen  by  lot  .from  5  Democrats  and 
4  Republicans.  What  is  the  probability  that  it  will  bo  composed 
of  2  Democrats  and  2  Republicans  ? 

11.  From  a  bag  containing  5  black  balls,  6  red  balls,  and  7 
white  balls,  4  balls  are  drawn  at  random.  What  is  the  probar 
bility  that  all  are  black  ?  that  2  are  black  and  2  are  white  ? 

12.  What  is  the  probability  that  6  can  be  thrown  in  a  single 
throw  of  two  dice  ? 

13.  What  is  the  chance  of  throwing  over  20  with  four  dice  ? 

14.  From  an  urn  containing  40  tickets,  numbered  from  1  to  40, 
four  tickets  are  drawn.  If  each  ticket  drawn  entitles  the  holder 
of  the  duplicate  to  a  prize  and  A  holds  duplicates  of  two  tickets, 
what  is  the  chance  of  A's  obtaining  at  least  one  prize  ? 

ADV.  ALG.  —  29 


450  PROBABILITY 

15.  A"  has  three  shares  in  a  lottery  that  ofPors  2  prizes  to  18 
blanks.  B  has  two  shares  in  a  lottery  that  offers  5  prizes  to  25 
blanks.      Compare  their  chances  of  drawing  at  least  one  prize. 

■    Suggestion.  — Since  out  of  C^  ways  of  drawing  3  tickets  from  20  there 
are  C  3*  ways  of  drawing  three  blanks,  the  probability  that  A  will  draw  .3 

blanks  is  -^  =  ^^  '  ^^  '  ^^'  ■=  —  ■     Hence,  Prin.  2,  the  probability  that  A  will 
(>|>     20 .  19  •  18     95  ^ 

fail  to  draw  .3  blanks,  and  therefore  draw  at  least  one  prize,  is  1  —  f f ,  or  ||, 

16.  A  has  three  tickets  in  a  lottery  containing  10  prizes  and  90 
blanks.  B  has  three  tickets  in  another  lottery  containing  5  prizes 
and  45  blanks.  Show  that  B  has  a  better  chance  than  A  of  win- 
ning at  least  one  prize. 

506.  When  two  or  more  events  are  regarded,  in  the  occurrence 
or  non-occurrence  of  some  or  all  of  them,  as  joining  to  make  one 
event,  the  resulting  event  is  called  a  Compound  Event. 

Thus,  if  an  ace  is  thrown  with  each  of  two  dice,  these  two  events  may  be 
regarded  as  joining  to  produce  a  compound  event,  namely,  throwing  a  pair  of 
aces.  • 

Again,  if  the  event  of  throwing  an  ace  with  one  die  is  joined  to  the  event 
of  failing  to  throw  an  ace  with  the  other  die,  the  occurrence  and  non-occur- 
rence of  the  event  of  throwing  an  ace  with  a  single  die  are  joined  to  produce 
a  compound  event,  namely,  throwing  only  one  ace  with  two  dice. 

507.  Events  that  compose  a  compound  event  are  said  to  be 
Independent  or  Dependent  according  as  the  occurrence  of  one  does 
not  or  does  affect  the  probability  of  the  occurrence  of  the  other  or 
others. 

Thus,  in  the  event  of  throwing  double  sixes  with  two  dice,  the  component 
events  are  independent.  But  in  the  compound  event  that  six  shall  be  thrown 
with  one  die  on  the  second  trial,  the  component  events  are  dependent,  since 
the  success  of  the  second  trial  depends  partially  upon  the  failure  of  the  first 
trial. 

508.  Events  that  form  a  compound  event  by  joining  the  occur- 
rence of  each  event  to  the  non-occurrence  of  each  of  the  others 
are  called  Mutually  Exclusive  Events. 

The  compound  event  of  throwing  over  9  with  two  dice  is  composed  of  three 
mutually  exclusive  events.     For  when  10  is  thrown,  neither  11  nor  12  can  be 


PROBABILITY  451 

thrown  ;  when  11  is  thrown,  neither  10  nor  12  can  be  thrown ;  and  when  12 
is  thrown,  neither  10  nor  11  can  be  thrown. 

509.  Principles.  —  3.  Tlie  probability  of  a  compound  event  com- 
posed of  independent  events  is  equal  to  the  product  of  the  probabilities 
of  the  independent  events. 

4.  If  p  is  the  probability  that  an  event  will  happen  in  one  trial, 
the  iwobability  that  it  ivill  happen  n  times  in  succession  is  p". 

5.  The  probability  that  all  of  two  or  more  independent  events  will 
fail  is  (1  -_pi)  (1  -p^  •••. 

Principle  3  may  be  established  as  follows : 

Let  P  be  the  probability  of  a  compound  event  composed  of  events  whose 
probabilities  are  pi,  p2i  Pz-,  •••• 

It  is  to  be  proved  that  P  =  piP2Ps  •••• 

Suppose  that  the  first  event  can  happen  in  a  ways  and  fail  in  b  ways,  all 
equally  likely  ;  and  that  the  second  can  happen  in  c  ways  and  fail  in 
d  ways,  all  equally  likely. 

Since  the  events  are  independent,  not  only  may  both  happen  together,  but 
either  may  happen  while  the  other  fails,  or  both  may  fail  together.  Hence, 
the  probability  that  both  will  happen  together  is  the  ratio  of  the  number  of 
ways  they  can  happen  together  to  the  whole  number  of  ways  they  can 
happen  or  fail. 

Since  each  of  the  a  cases  in  which  the  first  event  happens  may  be  asso- 
ciated with  each  of  the  c  cases  in  which  the  second  event  happens,  there  are 
ac  cases  in  which  both  can  happen  together. 

Similarly,  there  are  ad  cases  in  which  the  first  happens  while  the  second 
fails,  be  cases  in  which  the  first  fails  while  the  second  happens,  and  bd  cases 
in  which  both  fail  together. 

Hence,  if  the  probability  that  both  will  happen  together  is  p, 


ac  +  ad  +  be  +  bd 

n.  c 

■PiPa- 


a  +  b    c  + 

If  there  is  a  third  event,  whose  probability  of  happening  is  ps,  the  con- 
currence of  the  three  events  may  be  regarded  as  the  concurrence  of  two 
events  by  considering  the  concurrence  of  the  first  two  as  a  single  event  whose 
probability  of  happening  is  p  =  pip-2.  Hence,  the  probability  of  the  three 
events  happening  together  is  pps,  or  pip-iPs- 

Since  this  method  can  be  extended  to  the  consideration  of  any  number  of 
independent  events,  P  =  piPiPs  •••• 

The  proofs  of  Principles  4  and  5  are  left  for  the  student. 


452  PROBABILITY 

Problems 

1.  The  probability  that  A  will  die  within  20  years  is  -^,  and 
that  his  wife  will  die  during  that  interval  is  -^^.  What  is  the 
probability  that  at  the  end  of  20  years  (1)  both  will  be  dead? 
(2)  both  will  be  living  ?  (3)  A  will  be  living  and  his  wife  dead  ? 
(4"^  A  will  be  dead  and  his  wife  living  ? 

Solution.  —  1.    Prin.  3,  P  =  ^^  x  i\  =  ^V 

2.  Prin.  5,  i'^  (1  -  A)(l  -  A)=  A- 

3.  Prin.  5,  Prin.  3,  p  =  (1  _  jl)  ^3^  =  .^zi^, 

4.  Prin.  ii,  Prin.  3,  P  =  (1  -  t\)  A  =  \. 

2.  A  crosses  the  United  States,  passing  through  five  cities 
where  he  has  friends.  If  the  odds  are  3  to  1  against  his  meeting 
a  friend  in  any  one  of  the  five  cities,  what  is  the  probability  that 
he  will  meet  a  friend  in  each  city  ? 

510.  Principle  6.  —  The  prohahiUty  of  a  compound  event  com- 
posed of  a  series  of  events,  dejyendent  or  independent,  is  equal  to  the 
probability  that  the  first  happens,  multijilied  by  the  probability  that, 
after  the  first  has  happened,  the  second  happens,  etc. 

For  if  the  events  are  independent,  considering  them  to  happen  in  any 
order  does  not  modify  the  probability  of  any  of  them  ;  and  if  they  are 
dependent,  after  each  has  happened  and  so  has  modified  the  probability  of 
the  following  event,  the  modified  event  may  be  regarded  as  independent  of 
the  preceding  event. 

Problems 

1.  From  a  bag  containing  3  white  balls  and  5  black  balls  A 
draws  a  ball,  and  after  A  has  replaced  the  ball  B  draws  a  ball. 
What  is  the  probability  that  A  draws  a  white  ball  and  B  a  black 
ball  ?  What  would  have  been  the  probability  of  A  drawing  a 
white  ball,  and  B  a  black  ball,  had  A  not  replaced  the  ball  he 
drew? 

Solution 

Since  by  replacing  the  ball  A  removes  the  effect  of  his  drawing  upon  the 
probability  that  B  draws  a  black  ball,  the  events  under  the  first  supposition 
are  independent,  but  under  the  second  they  are  dependent. 

In  the  first  case,  since  the  separate  probabilities  are  |  and  f,  the  joint 
probability  is  |  x  f ,  or  ^|. 


PROBABILITY  453 

In  the  second  case,  supposing  that  A  draws  a  white  ball,  B's  cliance  of 
drawing  a  black  ball  out  of  2  white  balls  and  5  black  balls  is  f.  Tliat  is,  the 
joint  probability  of  A  drawing  a  white  ball  and  B  a  black  ball  would  have 
been  f  x  f ,  or  \\. 

2.  A  starts  on  an  ocean  voyage.  The  probability  that  the  ship 
will  encounter  a  storm  is  |.  If  she  does,  the  probability  that  she 
will  spring  a  leak  is  y'^,  but  10  to  1  if  she  springs  a  leak  her 
engines  will  be  able  to  pump  her  out.  If  they  fail,  the  proba- 
bility that  the  compartments  will  keep  her  afloat  is  f.  If  she 
sinks,  it  is  an  even  chance  that  any  one  passenger  will  be  saved 
by  passing  boats.  What  is  the  probability  that  A  will  be  lost 
at  sea  ? 

511.  Principle  7.  —  TJie  probability  of  a  compound  event  com- 
posed of  two  or  more  mutually  exclusive  events  is  equal  to  the  sum 
of  their  separate  p)Tobai)ilities. 

For  if  one  event  can  happen  a  times  out  of  n,  another  h  times  out  of  ra, 
another  c  times  out  of  n,  etc.,  if  whenever  one  event  happens  the  others  fail, 
such  a  union  of  occurrence  and  non-occurrence  can  happen  a  +  6  +  c  +  ••• 
times  out  of  n  times. 

Hence,  the  probability  of  such  a  compound  event  is 

or«  +  ^+^+...; 
n      n      n 

that  is,  when  it  is  impossible  for  more  than  one  of  a  series  of  events  to 
happen  at  the  same  time,  the  probability  that  some  one  of  them  will  happen 
is  the  sum  of  the  probabilities  of  the  individual  events. 

512.  Let  p  be  the  probability  that  an  event  will  happen  in  one 
trial,  and  let  it  be  required  to  find :  (a)  the  probability  that  it 
will  happen  exactly  r  times  in  n  trials ;  (6)  the  probability  that 
it  will  happen  at  least  r  times  in  n  trials. 

(a)  By  Prin,  4,  the  probability  that  the  event  will  happen  any 
particular  r  times  is  p'',  and  since,  Prin.  1,  the  probability  that 
the  event  will  fail  in  any  one  trial  is  1—p,  the  probability  that 
it  will  fail  the  remaining  n  —  r  times  is  (1  —  />)"~^ 

Hence,  Prin.  3,  the  probability  of  the  compound  event,  that  the 
event  will  happen  in  any  parficidar  r  trials  and  fail  in  the  remain- 
ing n  —  r  trials,  is  ^""(l  —  j?)""''. 


454 


PROBABILITY 


Since  out  of  n  trials,  exactly  r  successful  trials  can  be  made  in 
as  many  ways  as  there  are  combinations  of  r  things  out  of  n 
different  things,  when  each  excludes  the  others  the  probability 
that  the  event  will  happen  exactly  r  times  out  of  n  is  C"  times 
as  great  as  the  probability  that  the  event  will  happen  in  any 
particular  r  trials  and  fail  in  the  remaining  n  —  r  trials. 

Hence,  the  probability  that  the  event  will  happen  exactly  r  times  in 
n  trials  is 

CrP' {1  -  py-'.  (A) 

(6)  Happening  at  least  r  times  in  n  trials  includes  the  following 
mutually  exclusive  events,  namely,  happening  exactly  n,  n  —  1, 
'•',  r  times  in  n  trials.  Substituting  these  values  successively  for 
r  in  C"^'  (1  —  pY'",  and  adding,  Prin.  7,  the  probability  that  the 
event  will  happen  at  least  r  times  in  n  trials  is 

p"  +  Cfp"-^  (1  -  i?)  +  C?p"-2  (1  -  pY 

+  ■■'  +  c^p^(1-pY-% 


or 


j9"  +  np"-*(l  —p)  + 


n(n-l)    „_2^.  _ 


\1 


p"-\i-py 


+  •  •  •  to  w  —  r  +  1  terms. 


0') 


Miscellaneous  Problems 
1.    Show  that  the  probabilities  of  throwing  2,  3,  4,  •••,  12,  re- 
spectively, in  one  throw  of  2  dice  are  3V,  ^V  s%,  ^,  j\,  ^^,  i\^  ^% 


3%  3F»  -sV  respectively. 


Solution 


Since  a  die  has  six  faces  numbered  from  1  to  6,  the  probability  of  throw- 
ing any  particular  number  from  1  to  6  with  one  die  is  J,  and  the  probability 
of  throwing  the  same  number  with  two  dice,  or  of  throwing  any  two  desig- 
nated numbers,  one  with  each  die,  is  J  x  J,  or  ■^. 

2,  3,  4,  •••,  12  can  be  thrown  as  follows: 


2  = 

3  = 

4  = 


and  so  on. 


1+1, 
1+2  1 
2  -f  1  I ' 

14-3 

2  +  2 

3  +  1 


12: 
11 

10  = 


+ 


{5  +  6 
I  6  +  5 

f4  +  6 
5  +  6 
6+4 


I 


PROBABILITY  455 

Hence,  2  and  12  can  be  thrown  each  in  1  way  ;  3  and  11,  each  in  2  ways ; 
4  and  10,  eacli  in  3  ways  ;  5  and  9,  each  in  4  ways  ;  (5  and  8,  each  in  5  ways  ; 
and  7,  in  6  ways.  Therefore,  since  the  probability  of  throwing  two  dice  in 
any  designated  way  is  5*5,  if  po,  p.i,  •••,  denote  the  probabilities  of  throwing 
2,  3,  •••,  respectively,  in  one  throw, 

P-2  =  j\,   Ps  =  /s,  Pi  =  t%,   P5  =  3*S^   P6  =  A'  P^  =sV 

Ps  =  A>  P9  =  Ai  i'lo  =  3^S'  Pn  =  A'   P12  =  ih- 

2.  What  is  the  probability  of  throwing  an  ace  (a)  six  times  in 
six  trials  ?  (6)  at  least  once  in  six  trials  ? 

Solution 

(a)  Since  the  probability  of  throwing  an  ace  in  a  single  trial  is  |,  the 
probability  of  throwing  an  ace  six  times  in  succession  is  (J)^  =  jsiss- 

The  same  result  riiay  be  obtained  by  substituting  ^  for  p,  6  for  n,  and  6 
for  r  in  formula  (A). 

(6)  If  the  ace  is  not  thrown  at  least  once  in  six  trials,  then  the  person 
who  throws  the  die  must  fail  to  throw  the  ace  six  times  in  succession. 

Since  the  probability  of  failing  to  throw  an  ace  is  1  —  ^,  or  |,  with  each 
trial,  the  probability  of  failing  to  throw  an  ace  six  times  in  succession  is(|)^ 

Hence,  the  probability  of  throwing  an  ace  at  least  once  in  six  trials  is 
1  ~(l)^i  ^^  mH'  which  is  slightly  less  than  |. 

The  same  result  may  be  obtained  by  substituting  ^  for  p,  |  for  I  —  p, 
6  for  n,  and  1  for  r  in  formula  (B).  Thus,  if  the  probability  of  throwing  at 
least  one  ace  in  6  trials  is  P, 

/iy/5y      6-5-4-3-2nW5\s 
VOy   Uj        1-2.3.4.5U/UJ 


6.5.4.3/lW5\*  ,  6.5.4.3.2 


1.2.3.4 

=  a +  §)«-(!)«  =  1 -(!)«  =  HHi- 

3.  If  A  and  B  take  turns  in  the  order  named  in  drawing  one 
ball  from  a  bag  containing  2  white  balls  and  4  black  balls,  what 
chance  has  each  of  being  the  first  to  draw  a  white  ball  ?  If  A 
and  B  are  drawing  for  a  prize  of  $  25,  how  much  less  is  B's  chance 
worth  than  A's  ? 

Solution 

First  trial.  — The  probability  that  A  will  draw  a  white  ball  is  |,  or  ^,  and 
the  probability  that  he  will  draw  a  black  ball  and  so  give  B  an  opportunity 
to  draw  from  the  remaining  2  white  balls  and  3  black  balls  is  1  —  |,  or  |. 


456  PROBABILITY 

Second  trial.  —  The  probability  that  B  will  have  an  opportunity  to  draw 
and  will  draw  a  white  ball  is  §  x  §,  or  ^,  and  the  probability  that  he  will 
draw  a  black  ball  and  so  give  A  an  opportunity  to  draw  from  the  remaining 
2  white  balls  and  2  black  balls  is  |  (1  -  |),  or  f 

Third  trial. — The  probability  that  A  will  have  an  opportunity  to  draw 
and  will  draw  a  white  ball  is  §  x  ^,  or  \,  and  the  probability  that  he  will 
draw  a  black  ball  and  thus  give  B  an  opportunity  to  draw  from  the  remain- 
ing 2  white  balls  and  1  black  ball  is  |(1  —  ^),  or  ^. 

Fcnirth  trial.  — The  probability  that  B  will  have  an  opportunity  to  draw 
and  will  draw  a  white  ball  is  |  x  |,  or  t^^,  and  the  probability  that  he  will 
draw  a  black  ball  and  thus  give  A  an  opportunity  to  draw  from  the  remain- 
ing 2  white  balls  is  J  (1  —  |),  or  ^. 

Fifth  trial.  —  The  probability  that  A  will  have  an  opportunity  to  draw 
and  will  draw  a  white  ball  is  jij  x  f,  or  j^^,  and  the  probability  that  B  will 
have  another  opportunity  to  draw  is  JI5  x  S,  or  0. 

Hence,  A's  chance  for  the  prize  is  ^  +  i  +  1*5,  or  |,  and  B's  chance  is 
^  +  j^,  or  |.  Since  the  prize  is  worth  f  25,  A's  chance  is  worth  |  of  $  25, 
or  $  15,  and  B's  chance  is  worth  f  of  $25,  or  $  10,  that  is,  $5  less  than  A's. 

Note.  —  A's  expectation  is  $15  and  B's  expectation  is  $10. 

4.  What  is  the  probability  of  throwing  more  than  10  in  a 
single  throw  with  two  dice  ? 

5.  Compare  the  probabilities  of  throwing  3  with  one  die  and 
6  with  two  dice. 

6.  The  probabilities  of  three  events  are  |,  |,  and  \,  respec- 
tively. What  is  the  probability  that  one,  but  not  more  than  one 
of  them,  Avill  happen  ? 

7.  The  odds  in  favor  of  A's  solving  a  certain  problem  are  2 
to  1,  and  the  odds  in  favor  of  B's  solving  it  are  3  to  2.  If  both 
attempt  the  solution,  what  is  the  probability  that  it  will  be  solved  ? 
AVhat  is  the  probability  that  both  will  solve  it  ? 

8.  If  4  coins  are  tossed,  what  is  the  probability  that  at  least 
one  will  fall  head  ?  that  one  and  only  one  will  fall  head  ? 

9.  Which  is  the  greater,  the  probability  of  throwing  at  least 
one  ace  in  six  trials,  or  the  probability  of  throwing  at  least  one 
head  of  a  coin  in  two  trials  ? 

10.  Show  that  the  probability  of  throwing  an  ace  with  a  single 
die  is  ^  in  two  trials  and  ^^  in  thi-ee  trials. 


PROBABILITY  457 

11.  What  is  the  chance  that  a  backgammon  player  who  has 
two  throws  with  two  dice  will  make  exactly  10  ? 

12.  The  door  of  A's  house  has  a  spring  lock  that  fails  to  lock 
the  door  2  times  out  of  5.  A  comes  home,  and  in  the  darkness 
cannot  distinguish  between  6  keys,  one  of  which  fits  the  lock. 
What  is  the  probability  that  he  will  get  in  without  a  key  ?  with 
the  first  key  he  uses  ?  with  the  first  two  keys  ?  with  the  first 
three  keys  ? 

13.  What  is  the  chance  of  throwing  just  30  in  three  throws  of 
two  dice  ? 

14.  Find  the  probability  of  throwing  double  fives  with  two  dice 
exactly  3  times  out  of  4. 

15.  A,  B,  and  C  take  turns,  in  the  order  named,  in  drawing 
one  ball  from  a  bag  containing  2  black  balls  and  4  balls  of  other 
colors,  agreeing  that  the  first  to  draw  a  black  ball  shall  receive 
$  150.     What  is  the  value  of  the  expectation  of  each  ? 

16.  From  a  bag  containing  10  5-dollar  bills  and  20  2-dollar 
bills,  I  have  the  privilege  of  drawing  two  bills  at  random.  What 
is  the  value  of  my  expectation  ? 

17.  A,  B,  and  C  toss  a  coin,  in  the  order  named,  until  one  wins 
by  tossing  head.     Find  the  chance  that  each  has  of  winning. 

18.  A  and  B  throw  a  die  alternately,  in  the  order  named,  until 
one  wins  by  throwing  an  ace.     What  chance  has  each  of  winning  ? 

19.  A  and  B  toss  a  dollar  alternately,  in  the  order  named,  agree- 
ing that  the  dollar  shall  go  to  the  one  who  first  throws  head. 
What  is  the  value  of  the  expectation  of  each  ?  If  the  dollar  is 
their  joint  property,  each  owning  half,  what  is  the  probable  gain 
or  loss  of  each  ? 

20.  From  a  bag  containing  10  balls  numbered  1,  2,  3,  •••,  10, 
five  balls  are  drawn  at  random,  each  ball  being  replaced  before 
the  next  is  drawn.  Find  the  probability  of  drawing  the  ball 
numbered  4  exactly  three  times ;    at  least  twice. 

21.  A  machinist  works  300  days  a  year.  If  the  odds  are  1000 
to  1  against  his  meeting  with  an  accident  on  any  particular  work 
day,  show  that  the  odds  against  his  escaping  injury  for  5  years 
are  about  3  to  1.     (Use  logarithms.) 


SIMPLE    CONTINUED    FRACTIONS 


513.    An   expression    of    the   form   a-\ -r 

Continued  Fraction.  c  -\ 

e  + 

It  is  usually  written  in  the  more  compact  form 


is  called  a 


a  + 


c+  e-t- 


514.  A  continued  fraction  in  which  each  numerator  is  1  and 
each  denominator  is  a  positive  integer  is  called  a  Simple  Con- 
tinued Fraction. 

The  integral  part  a  may  be  either  a  positive  integer  or  zero. 

In  this  chapter  only  simple  continued  fractions  will  be  discussed,  and  for 
the  sake  of  brevity  these  will  be  called  continued  fractions. 


515.  A  continued  fraction  is  said  to  be  terminating  or  infinite 
according  as  the  number  of  denominators  is  finite  or  indefinitely 
great. 

A  terminating  continued  fraction  may  be  reduced  to  a  simple 
fraction  as  in  §  184. 

516.  In  the  continued  fraction 
111 


ttj  + 


02+    «3+    «4-| 

which  will  be  used  as  the  general  symbol  of  a  continued  fraction, 
Ui  is  called  the  first  convergent,  ai-\ —  the  second  convergent, 
tti  -1 the  third  convergent,  and  so  on. 


«£+  Os 


468 


SIMPLE   CONTINUED   FRACTIONS  459 

When  ttj  is  lacking,  0  is  regarded  as  the  first  convergent. 
«i)  <*2j  %)  ••■  s-re  called  partial  quotients. 

Og  -\ ,    a^-\ ,  •••  are  called  complete  quotients. 

a3  +  '"  «4-l 

517.    Let    -   be  a  fraction  whose  terms  are  integers  greater 
h 
than  1. 

Dividing  a  by  6,  let  a^  be  the  quotient  and  5i  the  remainder. 

r^,  a  5.  1 

Then,  -  =  a,  +-  =  a,  +  j- 

Dividing  h  by  6i,  let  ttj  be  the  quotient  and  h^  the  remainder. 

r^i  a  1  1 

Then,  r  =  «iH r  =  '^'i-l r' 

0  62  1 

h. 

Continuing  this  process  exactly  as  in  finding  the  H.  C.  D.  of  a 
and  h,  the  result  is  a  continued  fraction.  Since  a  and  h  are  posi- 
tive integers,  the  process  must  finally  come  to  an  end  when  the 
H.  C.  D.,  or  if  none  exists,  then  unity  itself,  is  obtained  as  a 
divisor.     Hence, 

A  fraction  tchose  terms  are  integers  greater  than  1  may  be  con- 
vetied  into  a  terminating  continued  fraction. 

Examples 

1.  Keduce  ^^  to  a  continued  fraction. 

Solution.  24J 1 51!^  =  ai 

144 

7_|24[3  =  a2 
21 
3_|7t2  =  a8 
6 

.    1^  =  6+  —  —  -  lJ3[3  =  a4 

*  ■    24  3+2+3' 

Reduce  to  continued  fractions : 

2.  H-  3.    H-  4.    If  5.     ■^.  6.    H-  7.    IH- 


46a  SIMPLE   CONTINUED   FRACTIONS 

518.   In  «,  +  A^  -4-^,  let  ^,   ^^  ^^  ••,  ^-,  ... 

be  the  successive  convergents  reduced  to  simple  fractions.     Then, 
Pi_'h.    "Pi _  <^2<^i  +  1 .   V^ _ a^iSL^i  +  1)  +  ^1  _ asPi  +  Pi . 

91  ~"  1  '      92  ~       a2  •  1      '      93  ~  «3a2  +  1  "   ^392  +  9i   ' 

j94  ^  a4[a3(a2ai  + 1)  +  «i]  +  (a2«i+l)  _  ^4^3  +  Pi 
g*  ~  ai{a3fi2  +  1)  +  a,  ~~  «493  +  92 

From  an  examination  of  the  third  and  fourth  convergents  it  is 
seen  that  they  are  formed  according  to  a  certain  law.  If  the  wth 
convergent  obeys  this  law, 

Pn  _anPn-l  +  Pn-2  /-^N 

9»~«n9n-l  +  9n-2' 

It  remains  to  be  shown  that  if  (1)  is  true  for  the  wth  convergent, 
it  will  be  true  for  the  (n  +  l)th  convergent. 

Since  the  (n  +  l)th  convergent  differs  from  the   nth  only  in 

having  a„  + instead  of  a,„ 

^n+l 

»       ,  {^-~^n )Pn-^+Pn-2 

Fn+l  \ <^n  +  l/ 

9:;7i~7      r^;  ' 

(«n  +  ^)9«-l  +  9«-2 

multiplying  both  terms  by  a„_^i  and  rearranging, 

^  «n+l(<^n A.-1  +  Pn-2)  +  Pn-1 
a„+l(«n9n-l  +  9»-2)+  9«-l 

"n+l9n  T^  9n-l 

which  has  the  same  form  as  (1),  n  +  1  taking  the  place  of  n. 

Hence,  if  (1)  is  true,  (2)  is  true ;  that  is,  if  the  law  of  forma- 
tion of  successive  convergents  revealed  in  the  third  and  fourth 
convergents  is  true  for  any  particular  convergent,  it  holds  true  for 
the  next  convergent. 

Therefore,  since  the  law  is  true  for  the  fourth  convergent,  it 
holds  for  the  fifth ;  and  being  true  for  the  fifth,  it  holds  for  the 
sixth ;  and  so  on. 

Hence,  formula  (1)  expresses  the  law  of  formation  of  the  suc- 
cessive convergents  to  a  continued  fraction,  each  convergent  after 


SIMPLE   CONTINUED  FRACTIONS  461 

the  second  being  obtained  from  the  two  convergents  immediately 
preceding  it. 

This  formula  is  called  the  recurrence  formula. 

Examples 

1.  Find  the  first  six  convergents  to 

3        1       1      1       1         1 

^2+6+  2+  6+  2+...' 

Solution.  — The  first  two  convergents,  3  and  |,  are  found  by  inspection, 
and  the  first  convergent  is  written  |  because  its  denominator  is  needed  in 
forming  the  denominator  of  the  third  convergent.    By  the  recurrence  formula, 

3d  convergent  =  ^AJ1±1  =  ^ ■       4th  convergent  =  2(45)_+I  =  21 
6(2)4-1      13  2(13)4-2      28 

5th  convergent  =  ^Jm±^  =  627    ^^^  eonvergent  =  2(627)+97  ^  1351 . 
6(28)4-13      181  2(181)4-28       390 

Hence,  the  first  six  convergents  are  3,  |,  f^,  ||,  ff|,  -VW- 
Note.  —  In  applying  the  recurrence  formula,  if  the  first  convergent  is  0, 
it  is  written  ^. 

Find  the  first  five  convergents  to 

2.  2+  —  —  -^^ ^ 4.  0-f—  ^ 


3.  1-f 


1  +  3+5+7  +  .. .  2+1  +  3+2  + 

1   1 1   1       g   111    1 

2+3+4+5  +  ...*      '  1+1  +  1+2  +  ...* 


519.  To  illustrate  the  principles  proved  in  the  succeeding  ar- 
ticles the  student  may  calculate  and  compare  the  convergents  of 
any  continued  fraction,  as 

■^        1       1       1       1       1         1 
+  1  +  2  +  3+  1  +  2  +  3  +  ...' 
The  convergents  of  this  continued  fraction  are  1,  2,  ^,  i^,  ^|, 
1^,  105^  ...^  or  1,  2,  1.66,  1.7,  1.6923,  1.6944,  1.6942,  .••,  approxi- 
mately. 

It  is  observed  that : 

(1)  If  any  tvro  consecutive  convergents  are  chosen,  the  even 
convergent  is  always  greater  than  the  odd  one. 

(2)  The  even  convergents  form  a  descending  series  and  the  odd 
convergents  form  an  ascending  series. 

(3)  The  difference  between  consecutive  convergents  grows  less 


462 


SIMPLE   CONTINUED  FRACTIONS 


1 


and  less,  and  the  even  and  odd  convergents  seem  to  approach  a 
common  limit.  Later  the  student  will  be  able  to  show  that  this 
limit  is  ^(6  +  V6),  an  incommensurable  number  and  the  actual 
value  of  the  continued  fraction. 

The   accompanying   diagram   shows   the   relative   size  of   the 
successive  convergents,  represented  by  the  dotted  lines. 


520.  Every  terminating  continued  fraction  has  a  single  definite 
value,  which  is  the  same  as  that  of  its  last  convergent. 

To  show  that  every  infinite  continued  fraction  has  a  single 
definite  value,  it  is  necessary  to  prove  that: 

Principle  1.  —  T7ie  difference  between  any  two  consecutive  con- 
vergents, as  the  (n-\-l)th  and  the  nth,  is  numerically  equal  to  the 
recijyrocal  of  the  product  of  their  denominators.  This  difference 
is  negative  when  the  minuend  is  an  odd  convergent,  and  positive 
when  the  minuend  is  an  even  convergent  ;  that  is, 

Pn+i     j)„^(-l)"+^ 

220  -  221  -  1 


Thus,  in  §  519, 


aud 


9n+l9»      ' 

a  _  £i  =  22  _ 
35      Qi      13 

17 
10 

i>6  _  P6  =  ^  _ 

ge      25      36 

22 
13 

hi9«-P«9«+i  =  (-!)' 


(1) 


13-10 

793  -  792 

36  .  13 


13-10 

1 
36-13" 


SIMPLE   CONTINUED  FRACTIONS  463 

The  above  principle  may  be  established  as  follows  : 

Suppose  that  the  principle  is  true  for  the  nXh  and  (m  —  l)tli  convergents. 

Then,  ^-^  =  ^",  or  p„g„_i  -  p„_,g,,  =  (-  1)»  (2) 

Since  by  the  recurrence  formula  j}„+i=a„+ip„+p,i_i  and  g'„+i  =  a„+ig'„+g„_i, 

Pn+\qH-Pnqn+l-((tn+lPn+Pn-l)qn-Pn(an+iqn  +  qn-l) 

canceling,  =  Pn-iqn  -  Pnqn-i 

by  (2),  =-(-!)»=(- 1)(- !)«=(- l)«+i. 

Dividing  by  ,„,„,.,  f-_|.=  (^\ 

qn+l        ?n  Qn+iqn 

Therefore,  if  (2)  is  true,  (1)  is  true  ;  that  is,  if  the  difference  between  the 

nth  and  the  (n  — l)th  convergents  is  found  by  Prin.  1,  the  difference  between 

the  (n  +  l)th  and  nth  convergents  is  found  by  Prin.  1. 

T,  4.    •         e -iQ  Pi     Pi     a2ai+l      ai         1 

But  since,  §  ol8,  - — ■^—  — 


gi  02  1        02  •  1 

the  principle  is  true  for  the  first  two  convergents.  Therefore,  it  holds  for 
the  second  and  third  ;  and  being  true  for  the  second  and  third,  it  holds 
for  the  third  and  fourth  ;  and  so  on.     Hence,  the  principle  is  true  universally. 

Corollary.*  —  The  convergents  to  a  continued  fraction  are 
fractions  in  their  loivest  terms. 

For  in  (1),  if  Pn  and  g„,  or  p„+i  and  qn+ii  had  a  common  factor,  this 
factor  would  be  a  factor  of  Pn+iqn  —Pnqn+h  or  of  ±  1,  which  is  impossible. 

521.  Principle  2.  —  TJie  even  convergents  to  a  continued  frac- 
tion form  a  descending  series  and  the  odd  convergents  an  ascending 
series. 

The  above  principle  may  be  established  as  follows : 

By  Prin.  1,  P^  _  Pn  ^  i^iy^  ^  _  tl^ ^  (^ 

g»+l        ?n  Qn+iqn  9n+l9n 

and  ^»_^^:.^l21".  (2) 

qn        qn-1        qnqn-1 

Adding  (1)  and  (2),       g^^  _  ^i^  (gn.i- g«-i)(- 1)".  (3^ 

*   ^    ^  qn+l         qn-1  qn+\qnqn-l 

By  the  recurrence  formula,  g„+i  >q„>  g„-i,  and  since  they  are  all  positive, 
the  sign  of  (3)  depends  only  upon  the  sign  of  ( —  1)". 

If  n  is  odd,  the  convergents  in  the  first  member  of  (.3)  are  even  con- 
vergents and,  (— 1)»  being  negative,  their  difference  is  negative;  that  is, 
the  even  convergents  form  a  descending  series. 

*  A  corollary  to  a  principle  is  a  subordinate  principle  easily  deduced  from 
the  main  principle  or  from  some  part  of  its  proof. 


464  SIMPLE   CONTINUED  FRACTIONS 

If  n  is  even,  the  convergents  in  tlie  first  member  of  (3)  ai-e  odd  con- 
vergents  and,  (—  1)"  being  positive,  their  difference  is  positive  ;  that  is,  the 
odd  convergents  form  an  ascending  series. 

522.  Principle  3.  —  Every  infinite  continued  fraction  has  a 
single  finite  value,  which  is  the  common  limit  of  the  terms  of  the 
descending  series  formed  by  its  even  convergents  and  the  terms  of 
the  ascending  series  formed  by  its  odd  convergents,  as  the  number 
of  convergents  increases  without  limit. 

The  above  principle  may  be  established  as  follows : 

By  Prin.  1  any  even  convergent  is  greater  tlian  the  preceding  odd  con- 
vergent, which  by  Prin.  2  is  greater  than  any  preceding  odd  convergent. 
Therefore,  every  even  convergent  is  greater  than  every  odd  convergent  that 
precedes  it. 

By  Prin.  2  any  even  convergent  is  greater  tlian  any  following  even  con- 
vergent, and  by  Prin.  1  each  of  the  latter  is  greater  than  either  of  the  adja- 
cent odd  convergents.  Therefore,  every  even  convergent  is  greater  than 
every  odd  convergent  that  follows  it. 

Hence,  every  even  convergent  is  greater  than  every  odd  convergent. 

It  is  evident  from  the  law  of  formation  of  the  denominators  of  successive 
convergents  that  they  form  an  ascending  series,  in  which  qi  <q2<Q3  <  •••• 

Therefore,  in        i^^,    (^ii^,  ..,   t^^,    (-1)""\  .., 

9251  QsQ-i  qnqn-i      qn+iqn 

whose  terms,  Prin.  1,  are  the  differences  of  successive  convergents,  one  even 
and  the  other  odd,  as  n  increases  without  limit  the  nth  difference  approaches 
zero  as  a  limit. 

Since  every  even  convergent  is  greater  than  every  odd  convergent,  and  as 
the  number  of  convergents  taken  increases  without  limit,  the  difference 
between  successive  convergents  approaches  zero  as  a  limit,  the  terms  of  the 
descending  series  of  even  convergents  and  the  terms  of  the  ascending  series  of 
odd  convergents  approach  a  common  limit,  which  is  the  value  of  the  continued 
fraction  itself. 

Corollary. —  The  value  of  an  infinite  continued  fraction  lies 
between  any  two  consecutive  convergents. 

523.  Principle  4. — Ayiy  convergent  to  an  infinite  continued 
fraction  is  nearer  the  continued  fraction  than  any  preceding 
convergent. 

The  above  principle  may  be  established  as  follows : 

Let  X  represent  the  value  of  the  continued  fraction. 

To  show  that  the  rath  convergent  is  nearer  to  x  than  the  (n  —  l)th  con 
vergent,  first  find  x  in  terms  of  these  convergents. 


I  SIMPLE   CONTINUED   FRACTIONS  465 

By  the  recurrence  formula  the  (n  +  l)th  convergent  involves  the  nth 
and  the  {n  —  l)lh  convergents  and  the  (n  +  \)i\\  partial  quotient,  an+\.  Let 
k  represent  the  complete  {n  +  l)th  quotient.  Since  when  the  complete  quo- 
tient is  substituted  for  the  partial  quotient  the  result  is  the  fraction  itself, 

kPn  +  Pn-l 


Therefore,*    x  ~ 


kqn  +  qn-i 

Pn-l        k(Pnqn-l  ~  Pn-lQn)  k 


Qn  1        qn-i{lcq„  +  qn-i)        qn-i(kq„  +  g„_i) 


and  ^~a;  = 


Pn       ^       JPn?n-l~Pn-l9n 


<?n  qn{kq„  +  g„_i)       q„(kqn  +  q„-i) 

Since  q„>qn-i  and  A;>1,  the  difference  between  the  nth  convergent 
and  X  is  less  than  the  difference  between  the  (n  —  l)th  convergent  and  x ; 
that  is,  any  convergent  is  nearer  to  the  continued  fraction  than  the  next  pre- 
ceding convergent,  and  is  nearer,  therefore,  than  any  preceding  convergent. 

524.  Principle  5.  —  Any  convergent  to  a  continued  fraction  is 
nearer  to  the  value  of  the  continued  fraction  than  any  other  fraction 
whose  denominator  is  smaller  than  that  of  the  convergent. 

The  above  principle  may  be  established  as  follows  : 

Let  X  be  the  value  of  a  continued  fraction,  ^  its  nth  convergent,  and  ? 

qn  b 

any  other  fraction  whose  terms  are  positive  integers,  and  whose  denominator 
is  less  than  q„. 

It  is  to  be  proved  that  ^  is  nearer  to  x  than  _• 

a  Qn  Pn        ^  a 

If  possible,  let  t  be  nearer  to  x  than  — '     Then,  Prin.  4,  t  is  nearer  to 

X  than  ^5zl.    Therefore,  since  by  Prin.  3,  Cor. ,  x  lies  between  the  nth  and 

the  (n  —  l)th  convergents,  t  lies  between  them. 

Hence,  «     ^<^',.^\ 

b         Qn    \        qn        ?n-l 

or,  Prin.  1,  aq.^-bpn-^_J_^ 

bqn-l  qnQn-l  ^    ' 

Since  g„>  6,  making  the  denominators  in  (1)  alike  by  substituting  qn  for 
6  in  the  first  denominator  decreases  the  first  fraction. 

Hence,  in  order  that  -  may  be  nearer  to  x  than  — ,  we  must  have 

ag„_i~  &;?„  i<l, 

which  is  impossible,  because  a,  qn-\i  b,  and  p„_i  are  integers. 

*  The  sign  ~,  read  '  the  difference  between,'  indicates  that  the  less  num- 
ber is  to  be  subtracted  from  the  greater. 

ADV.  ALG.  — 30 


466  SIMPLE   CONTINUED  FRACTIONS 

Hence,  since  6  <  <?„,  t  is  not  nearer  to  x  tlian  —  ;  that  is,  the  ntli  conver- 

gent  is  nearer  to  x  than  any  other  fraction  whose  denominator  is  smaller 
than  that  of  the  nth  convergent. 

525.   Limit  of  error  of  the  nth  convergent. 

Since,  Prin.  3,  Cor.,  the  value  of  a  continued  fraction  lies  between 
the  values  of  any  two  consecutive  convergents,  the  error  made  by 
taking  the  nth  convergent  instead  of  the  whole  fraction  is  less 
than  the  difference  between  the  nth.  convergent  and  the  next  con- 
vergent. By  Prin.  1  this  difference  is  numerically  equal  to  the 
reciprocal  of  the  product  of  their  denominators.     Hence, 

Principle  6.  —  The  error  made  by  taking  the  nth  convergent  of 
an  infinite  continued  fraction  is  less  than  unity  divided  by  the  prod- 
uct of  the  denominators  of  the  nth  and  (n  -f  l)th  convergents. 

Since  error  < , 

<lnQn+l 

and  =  — ; ■ r< 


Qngn+i      9«(an+i7»  +  9„-i)      a„+i9„ 
the  error  corresponding  to  the  nth.  convergent  is  small  when  a„+i, 
the  next  partial  quotient,  is  large.     Hence, 

Corollary.  —  Any  convergent  that  immediately  precedes  a  large 
partial  quotient  is  a  near  approximation  to  the  value  of  the  continued 
fraction. 

Examples 

111111         1 


1.  Compute  the  value  of  10  -|- 


1  +  4+2  +  4+1  +  20+1  +  — 

correct  to  the  nearest  fourth  decimal  place. 

e  ^  .      10     11     54     119     530    649      (  ) 

Solution. — Convergents,  — ,    — ,    — ,   — ,    — •,   -—,   -Hr;,  •••• 
^  1       1       5       11       49      60      1249 

Since  the  error  must  be  less  than  \  of  .0001,  or  less  than  25^5,  by 
Prin.  6  the  first  convergent  that  is  sufficiently  accurate  is  -^,  whose  denomi- 
nator multiplied  by  the  next  denominator  exceeds  20000. 

Hence,  continued  fraction  =  ^  =  10.8167,  to  the  nearest  ten-thousandth. 

2.  Find  the  value  of  tt  =  3  +  ^  -^  -^  ^  ^  to  the 
nearest  sixth  decimal  place.                                    "          '   "" 

3.  If  2.20462  pounds  make  1  kilogram,  find  six  commensurable 
fractions  that  represent  approximately  the  ratio  of  a  kilogram  to 
a  pound,  and  estimate  the  error  for  each  fraction. 


SIMPLE   CONTINUED  FRACTIONS  467 

526.   To  convert  a  quadratic  surd  into  a  continued  fraction. 

Let  it  be  required  to  convert  2  +  Vll  into  a  continued  fraction. 
The  largest  integer  in  2  +  Vll  is  2  +  3,  or  5. 

2  +  Vll  =  5+Vll-3  =  5  +  ^^^~^ 

2 
rationalizing  the  numerator,  =  5  H =r 

VlH-3 

reducing  the  numerator  to  1,  =  5  H -:=n (1) 

Vll  +  3  ' 

2 
Similarly,  Vn+3  =  3  +  Vn^  =  3  +  ^;  (2) 

1 

also,         — 3—='+— I— =^+vnTi-  ^'^ 

2 
The  last  denominator  in  (3)  may  be  transformed  by  (2),  then 
the  last  denominator  in  the  result  may  be  transformed  by  (3),  and 
so  on.     Substituting  the  results  (2),  (3),  and  so  on  in  (1), 


or 


2+vn=5+-^^ — - 

3+       1 
•  Vll+3 

1 

3+-^ 
6+— 4— 

vn+3 

2 

^+    1 

3+1    , 

6+  1 

6+1      , 

3+    ^1 
Vll+3 

1 

«-6  +  ... 

o  1  ^/TT-^  \     ^      ^      ^ 

1 

^^^^^-^^3+6+3+6+...' 

in  which  the  partial  quotients,  3  and  6,  recur.     Such  a  continued 
fraction  is  called  a  periodic,  or  a  recurrimj  continued  fraction. 

By  the  method  just  exemplified,  any  positive  quadratic  surd 
may  be  converted  into  a  periodic  continued  fraction. 


468  SIMPLE   CONTINUED  FRACTIONS 

527.    To  convert  a  periodic  continued  fraction  into  a  quadratic  surd. 

1.    Let  it  be  required  to  convert  6  +  - —  - — into  a 

*      o+6+3+6-f--" 

♦ 

quadratic  surd.     The  recurring  part  is  marked  by  asterisks. 

1  1 

Then,  a;  =  6  H z z —  =  6  + 


3  +  i^"        3fl  (2) 

G  +  o  +  •  •  •  X 

Simplifying,  etc.,  a^  —  6a;  —  2  =  0,  a  quadratic  equation  whose 
roots,  3  +  Vil  and  3  —  VH,  are  quadratic  surds,  one  positive  and 
one  negative.  Since  the  given  continued  fraction  is  positive,  its 
value  X  is  the  positive  root,  3  +  VH. 

Note.  — The  second  member  of  (2),  being  a  terminating  continued  frac- 
tion, may  be  simplified  by  finding  its  third  convergent. 

2.    To  convert  5  + into  a  quadratic  surd, 

3+  6+  3+  6h 

let  x  =  5-\-  - — 


3+  6+  3+  6+--- 
1  hen,  X  —  5  =  ^ 


3+g        1         1  3+6  +  (x-5) 

"^3+6  +  ... 
Simplifying,  etc.,  ar^  —  4  a;  —  7  =  0,  whence  a;  =  2  +  VH. 

3.    To  convert  5  -\ —  - — into  a  quadratic  surd, 

4+  6+  3+  6+"- 
«  * 

first  find  the  value  of  the  complete  quotient  whose  integral  part 

is  the  first  recurring  partial  quotient  6. 

Byl,  6  +  ^^;^  =  «^  =  3+VlT.  (1) 

Next,  put  y  =  5  +  — -^^ —  =  5  +  —  --  (2) 

'^      ^         ^4+6+3+6  +  ---  4+a;  ^^ 

*  * 

Substituting  (1)  in  (2),  ?/  =  -f  (40  -  Vll). 

Hence,  every  periodic  continued  fraction  is  equal  to  a  quadratic 
surd  and  is  a  positive  root  of  a  quadratic  equation  with  rational 
coefficients. 


SIMPLE   CONTINUED  FRACTIONS  469 


The  proof 

is  as  follows : 

Let 

2/  =  6i+-^ 

^2  +  • 

111            11 

•  bn  +  ffll  +   a2  +  •••  Or  +  Ol  +  — 

--^.. 

1        1 

•  6„  +  X 

(1) 

In  (1), 

a-2  +  • 

=  aiH 

(2) 

Since  x  equals  the  (»•  +  l)th  convergent  of  (2),  by  the  recurrence  formula, 

xqr  +  qr-1 

Clearing  (3) ,  qrX^  -(Pr-  qr-i)x  -Pr-i  =  0, 

a  quadratic  in  x  whose  positive  real  root  (§  304,  I)  is 


2qr  ^  ^ 

Since  ai  recurs,  ai  must  be  equal  to  or  greater  than  1 . 
•  ■•  Pr^qr^ qr-i,  Pr  —  qr-1  IS  a  positive  integer,  and  x  is  real. 
To  prove  that  the  radical  in  (4)  is  a  quadratic  surd,  write 


VCiJr  -  qy-l)'^  +  4  Pr-\qr  =  ^(Pr  +  qr-l)'^  +  4:(P>-i<lr  -  Prqr-l) 


Prin.  1,  formula  (1),  =  VCp--  +  Qr-i)'^  ±  4. 

Since  no  positive  integral  square  increased  or  diminished  by  4  is  a  perfect 
square,  the  radical  is  a  quadratic  surd. 

Hence,  x  is  a  quadratic  surd. 

Therefore,  if  the  value  of  x  is  substituted  for  x  in  (1)  and  the  result  is 
simplified  by  repeated  rationalization  of  denominators  and  reduction  to  higher 
terms,  y  will  finally  be  reduced  to  the  form  of  a  quadratic  surd. 

Examples 

Reduce  each  of  the  following  to  a  periodic  continued  fraction 
and  find  a  near  approximation  (§  525)  to  its  value  : 

1.  Vl5.  3.   V7.  5.   V'2i.  7.  VlO. 

2.  Vi26.  4.  2  4-V7.  6.   Vl9.  8.  1 :  VlO. 

Reduce  the  following  to  quadratic  surds  : 
9.2  +  ^—^.  11.14-^      ^        ^ 


8  +  2  +  ...  3  +  2  +  2  +  ... 

«  »       # 

10.  1+-1-Tx-^^ 12.3+-^      111 


10  +  10  +  ...  1+1+1+1+6+1  + 


THEORY    OF   NUMBERS 


SCALES  OF  NOTATION 

528.  The  theory  of  numbers  treats  of  the  properties  of  positive 
integers.  In  this  chapter  the  word  number  will  be  used  in  the 
sense  of  positive  integer,  and  the  terms  integer,  even  number,  odd 
number,  power,  root,  etc.,  will  be  employed  in  the  same  sense  as 
in  arithmetic,  unless  the  contrary  is  stated. 

529.  The  Arabic  system  of  notation  is  called  the  decimal 
system,  because  ten  units  of  any  order  make  one  unit  of  the  next 
higher  order.     10  is  called  the  radix  of  the  scale  of  notation. 

Any  number  in  the  decimal  system  may  be  expressed  in  a  scale 
of  descending  powers  of  10,  the  exponent  of  the  highest  power 
being  one  less  than  the  number  of  digits  of  the  number. 

Thus,       256473  =  200000  +  50000  +  6000  +  400  +  70  +  3 

=  2(10)5  +  5(10)*  +  6(10)8  +  4(10)2  +  7(-io)  +  3. 

530.  Any  number  greater  than  1  may  be  used  as  the  radix  of  a 
scale  of  notation. 

The  meaning  of  the  expression  256473  in  the  decimal  scale  has  just  been 
shown.     But  in  the  scale  whose  radix  is  8, 

256473  =  2(8)5  +  5(^)4  +  6(8)8  +  4(8)2  ^  7(8)  +  3. 

In  the  binary  scale  the  radix  is  2  ;  in  the  ternary,  3 ;  in  the 
qimternaiy,  4  ;  in  the  quinary,  5 ;  in  the  senary,  6  ;  in  the  septen- 
ary, 7 ;  in  the  octary,  8 ;  in  the  nonary,  9 ;  in  the  denary,  or  deci- 
tnal,  10 ;  in  the  undenary,  11 ;  in  the  duodenary,  or  duodecimal,  12 ; 
and  so  on. 

The  general  symbol  for  any  radix  is  r. 

470 


I  THEORY  OF  NUMBERS  471 

531.  Principle. — Any  integer  N  may  he  expressed  in  the  scale 
ivhose  radix  is  r  in  one  and  only  one  loay. 

The  above  principle  may  be  established  as  follows : 

Divide  N  by  r,  and  let  the  quotient  be  Ni  and  the  remainder  s. 

Then,  N  =  Niv  +  s.  (1) 

Divide  Ni  by  r,  and  let  the  quotient  be  N^  and  the  remainder  q. 

Then,  Ni  =  N^r  +  g, 

whence,  by  (1),  iV  =  N^r"^  +  qr -\-  s.  (2) 

Continuing  the  process,  a  quotient  less  than  r  is  finally  obtained.  Suppose 
this  occurs  after  r  —  1  divisions,  and  denote  the  last  quotient  and  remainder 
by  a  and  6,  respectively. 

Then,  iV"  =  ar»-i  +  &r»-2  + \-qr+s. 

The  coefficients  a,  ft,  •••,  5',  s  are  the  digits  of  the  number,  and  since  each  is 
a  remainder  of  some  division  by  r,  each  is  less  than  r.  Also,  since  there  can 
be  but  one  remainder  less  than  r  for  each  division,  there  is  but  one  way  of 
expressing  N  in  the  scale  of  r. 

Corollaries.  —  1.  The  exponent  of  the  highest  power  of  the 
radix  is  one  less  than  the  number  of  digits. 

2.  In  any  scale  the  number  of  different  characters,  including  0, 
that  are  required  to  express  all  numbers  is  equal  to  the  radix. 

For  example,  in  the  duodecimal  scale  the  ten  Arabic  figures  and  two  other 
characters,  t  for  ten  and  e  for  eleven ,  are  required  ;  in  the  quinary  scale  only 
the  figures  0,  1,  2,  3,  4  are  required. 

532.  To  change  from  the  decimal  to  another  scale. 

Examples 
1.   Change  3010  from  the  decimal  to  the  senary  scale. 

Explanation.  —  Since  6  units  of  any  order  make  one 

PROCESS  ^"^^^  ^^  ^^^  next  higher  order,  3010  units  of  the  first 

order  make  501  units  of  the  second  order  and  4  units  of 

6  [3010  the  first  order ;  501  units  of  the  second  order  make  83 

6  [501  ...  4        units  of  the  third  order  and  3  units  of  the  second  order ; 

6  [83  ...3        and  so  on. 

^  M  g        p.  Hence,  3010  when  expressed  in  the  senary  scale  is  the 

^~  "'  sum  of  2  units  of  the  fifth  order,  1  unit  of  the  fourth 

Z  •••  i.        order,  5  units  of  the  third  order,  3  units  of  the  second 

order,  and  4  units  of  the  first  order. 

For  convenience  the  radix  of  the  scale  is  indicated  by  a  subscript  figure, 

which  may  be  omitted  in  the  decimal  scale.    Thus,  3010  =  215346. 


_6 
83 


472  THEORY  OF  NUMBERS 

2.  Express  in  the  octary  scale  50,  128,  and  5283. 

3.  Express  in  the  quinary  scale  12,  342,  and  6627. 

4.  Express  in  the  duodecimal  scale  15,  100,  and  6053. 

5.  Express  in  the  binary  scale  the  numbers  from  1  to  10. 

533.  To  change  from  any  scale  to  the  decimal  scale. 

Examples 

1.  Change  215346  to  the  decimal  scale. 

PROCESS 

21534  Explanation.  —  Since  each  unit  of  any  order  is  equal  to  6 

6  units  of  tiie  next  lower  order,  2  units  of  the  fifth  order  are  equal 

To  to  12  units  of  the  fourth  order,  and  adding  1,  the  number  of 

units  of  the  fourth  order  in  215346,  the  whole  number  of  units 

of  the  fourth  order  is  13. 

Continuing  in  a  similar  way  to  reduce  the  units  of  each 

6  order  to  units  of  the  next  lower  order,  the  whole  number  of 

601  units  of  the  lowest  order  is  found  to  be  3010,  which  is  expressed 

g  in  the  decimal  scale. 

3010 

Change  the  following  to  the  decimal  scale : 

2.  42;.         3.    66548.         4.   UQ^-         5.    6e4i2.         6.    IIIIOOI2. 

534.  Arithmetical  processes  in  any  scale. 

The  processes  are  performed  in  the  same  manner  as  in  the  deci> 
mal  scale.  The  student  must  simply  bear  in  mind  each  time  the 
number  of  units  of  each  order  required  to  make  one  of  the  next 
higher  order. 

When  the  process  is  complicated  or  when  different  scales  are  involved,  j.11 
the  numbers  may  first  be  reduced  to  the  decimal  scale. 

Examples 
Perform  the  operations  indicated : 

1.  38,2  +  45i2  +  e6i2.  3.   42415-33235.         5.    I3047--257. 

2.  IOI2  +  IIO2  +  lOllOlo.    4.   4e5823i2  x  15i2.       6.    Vi32i;. 

7.  Multiply  2I12  by  13^  and  express  the  product  in  the  decimal 
scale. 


THEORY  OF  NUMBERS  ,     473 

8.  Show  that  in  any  scale  the  expression  121  represents  a  per- 
fect square. 

9.  In  what  scale  is  5  times  6  expressed  by  36  ? 

10.  In  what  scale  is  \  of  the  number  100  equal  to  the  number  30  ? 

11.  Which  of  the  weights  of  1,  2,  4,  8,  •••  pounds  must  be  taken 
to  aggregate  75  pounds,  if  not  more  than  one  of  each  is  used  ? 

PRINCIPLES  OF  DIVISIBILITY 

535.  In  this  subject  '  divisible '  is  used  for  '  exactly  divisible.' 
Also,  since  there  is  no  remainder  when  zero  is  divided  by  any 
number,  zero  is  regarded  as  exactly  divisible  by  any  number. 

536.  If   the    number    JV^=  ar"^  +  &/•"--  -\ \-pr  +  qr  +  .s   is 

divided  by  r,  the  remainder  is  s,  the  last  digit ;  if  iV  is  divided  by 
9-^,  the  remainder  is  the  number  that  is  expressed  by  the  last  two 
digits ;  if  A^  is  divided  by  ?-^,  the  remainder  is  the  number  that  is 
expressed  by  the  last  three  digits ;  and  so  on. 

Likewise,  if  N  is  divided  by  a  factor  of  r,  the  remainder  is  the 
same  as  that  obtained  by  dividing  the  last  digit  by  that  factor ;  if 
N  is  divided  by  a  factor  of  r^,  the  remainder  is  the  same  as  that 
obtained  by  dividing  qr  +  s,  the  number  expressed  by  the  last  two 
digits,  by  that  factor  ;  and  so  on.     Hence, 

Principle  1.  —  (a)  If  a  number  in  the  scale  of  r  is  divided  hy 
y™,  the  remainder  is  the  number  that  is  expressed  by  the  last  m  digits 
of  the  given  riumber. 

(b)  If  the  number  is  divided  by  a  factor  of  r™,  the  remainder  is 
the  same  as  that  obtained  by  dividing  by  that  factor  the  number  that 
is  expressed  by  the  last  m  digits. 

Corollaries.  —  1.  In  the  decimal  system  a  number  ending  in  m 
ciphers  is  divisible  by  10"*. 

2.  In  the  decimal  system 

(a)  Every  number  ending  in  0  or  5  is  divisible  by  5. 

(b)  Every  mimber  ending  in  00,  25,  50,  or  75  is  divisible  by  25. 

(c)  Every  number  ending  in  three  ciphers  or  in  three  digits  that 
express  a  multiple  of  125  is  divisible  by  125.     And  so  on. 

3.  In  the  decimal  system 

(a)  Every  mimber  ending  in  0,  2,  4,  6,  or  8  is  divisible  by  2. 


474      ,  THEORY  OF  NUMBERS 

(b)  Every  number  ending  in  two  ciphers  or  in  two  digits  that 
express  a  multiple  of  4  is  divisible  by  4. 

(c)  Eve^-y  number  ending  in  three  ciphers  or  in  three  digits  that 
express  a  multiple  of  8  is  divisible  by  8.     And  so  on. 

Note,  —  If  the  radix  is  odd,  the  last  digit  of  a  number  does  not  indicate 
whether  the  number  is  even  or  odd.     Thus,  llg  =  6  ;  I25  =  7  ;  II25  =  32. 

537.  From     a?-»-^  -f-6r"-2  -\ ypt^  -\-qr    +s 

subtracting        a 4-_6 + f-p +  q      -j-  s 

the  result       =  a^-"-^  -  l)  +  6(r»-2-l)  -\ \-p{r^  -  1)  +  qO'-I), 

which,  §  111,  is  divisible  by  r  —  1.     Hence, 

Principle  2.  —  The  difference  betiveen  any  number  and  the  sum 
of  its  digits  is  divisible  by  the  radix  less  one. 

Corollary.  —  In  the  decimal  system  the  difference  between  any 
number  and  the  sum  of  its  digits  is  divisible  by  9. 

538.  Since  the  difference  between  any  number  and  the  sum  of 
its  digits  is  a  multiple  of  r  —  1,  r  being  the  radix,  and  since  the 
number  that  must  be  added  to  this  multiple  of  r  —  1  to  produce 
the  given  number  is  the  sum  of  the  digits,  it  follows  that : 

Principle  3.  —  (a)  TJie  remainder  obtained  by  dividing  a  mim- 
ber  in  the  scale  ofrbyr  —  1  is  the  same  as  that  obtained  by  dividing 
the  sum  of  the  digits  by  r  —  1. 

(b)  TJie  remainder  obtained  by  dividing  the  number  by  any  factor 
ofr  —  1  is  the  same  as  that  obtained  by  dividing  the  sum  of  the  digits 
by  that  factor. 

Corollaries. — 1.  A  number  in  the  scale  of  r  is  divisible  by 
r  —  1,  if  the  sum  of  its  digits  is  divisible  by  r  —  1;  or  by  any  factor 
of  r  —  1,  if  the  sum  of  the  digits  is  divisible  by  that  factor. 

2.   In  the  decimal  system 

(a)  Any  number  is  divisible  by  9,  if  the  sum  of  its  digits  is  divisible 
by  9. 

(b)  Any  number  is  divisible  by  3,  if  the  sum  of  its  digits  is  divisible 
by  3. 

(c)  Any  even  number  is  divisible  by  6,  if  the  sum  of  its  digits  is 
divisible  by  3. 

The  test  called  casting  out  the  nines  is  an  application  of  Prin.  3.  For 
applications  of  this  test,  see  the  author's  Standard  Arithmetic,  page  400. 


THEORY  OF  NUMBERS  475 

539.  Let  N  =  a?-""^  -\ \-hr*  +  hi^  +pr^  +  qr  +  s. 

Subtracting  the  coefficients  of  the  even  powers  of  r  and  also  the 
coefficients  of  the  odd  powers  of  r  with  their  signs  changed,  and 
calling  the  subtrahend  d, 

N=s  +  qr  +pi^  -{-kr^  -{-hr^  -\ 

d=  s—  q +j3 —  k +  h 

N-d=        5(r +  l)  +  ^(7-2-l)+ fc(j'3  + 1)4- /i(r*  _!)+.. . 

By  §  111,  this  result  is  divisible  by  r  +  1. 

Then,  let  JV—  d  =  m  (r  + 1),  m  being  an  integer. 

.-.  ]Sf=m(r  +  l)-\-d. 

Therefore,  if  d  is  divisible  by  r  +  1,  ^  is  also.     Hence, 

Prijj ciPLE  4.  —  A  number  in  the  scale  of  r  is  divisible  by  r  -\-l, 
if  the  difference  between  the  su7ns  of  its  alternate  digits  is  divisible 
by  r  + 1. 

Corollary.  —  In  the  decimal  system  ayiy  number  is  divisible  by 

11,  if  the  difference  bettoeen  the  sums  of  its  alternate  digits  is  divisible 
by  11. 

Examples 

Which  of  the  numbers  2,  4,  8,  16,  5,  10,  15,  20,  25,  100,  3,  6,  9, 

12,  11,  22,  33  are  divisors  of  the  following  ? 

1.  1000.  4.    6250.  7.    522.  10.    54945. 

2.  1001.  5.    1620.  8.    275.  11.    10240. 

3.  1012.  6.    2673.  9.    729.  12.   80080. 

PRIME  AND  COMPOSITE  NUMBERS 

540.  Suppose  that  x  is  prime  to  a  but  is  a  factor  of  ab. 

Since  -  can  be  converted  into  a  continued  fraction  that  termi- 

X 

nates  with  the  nth  convergent,  if  the  preceding  convergent  is  -, 

§  520,  (1),  «  ^-P  =  1,  or  aq-^px^  1. 

X      q      qx 

.'.  abq-^bpx  —  b. 

Since  ab  and  x  are  each  divisible  by  x,  the  difference  between 
ab  •  q  and  x  •  bp,  which  is  b,  is  divisible  by  x.     Hence, 


476  THEORY  OF  NUMBERS 

Principle  1.  —  Any  factor  of  ah  that  is  prime  to  a  must  be  a 
factor  of  b. 

Corollaries.  —  1.  If  x  is  prime  to  a,  b,  c,  •••,  it  is  prime  to 
their  product. 

2.  If  a',  b',  c',  •••  a»-e  pnme  to  each  of  the  numbers  a,  b,  c,  •••, 
then  a'b'c'  •••  is  prime  to  abc  •••. 

3.  Powers  of  different  prime  numbers  or  of  mimbers  prime  to 
each  other  are  prime  to  each  other. 

541.  By  definition,  every  composite  number  can  be  resolved 
into  at  least  two  factors,  each  greater  than  1  and  less  than  the 
number.  If  any  of  these  factors  is  composite,  it  may  be  resolved 
into  factors ;  and  if  this  process  is  continued  far  enough,  the  given 
number  will  finally  be  resolved  into  prime  factors. 

To  discover  whether  a  composite  number  has  more  than  one  set 
of  prime  factors,  suppose  N=  abc  •••,  in  which  a,b,c,  •••  are  prime 
numbers;  and  if  possible,  suppose  N=ABC  •  ••,  in  which  A,B,C,-" 
are  prime  numbers. 

Then,  abc--  =  ABC--. 

Since  A  must  divide  the  product  abc  •••,  and  since  a,  b,  c,  ••• 
are  prime  numbers,  A  must  be  equal  to  some  one  of  the  factors 
a,  b,  c,  ••-.  Similarly,  each  of  the  factors  B,  C,  •••  must  be  equal 
to  some  one  of  the  factors  a,  b,  c,  •••;  that  is,  each  factor  in 
ABC  '■■  corresponds  to  an  equal  factor  in  abc  •••.     Hence, 

Principle  2.  —  A  composite  number  may  be  resolved  into  one 
and  only  one  set  of  prime  factors. 

Corollary.  —  Every  composite  number  has  at  least  one  factor 
equal  to  or  less  than  its  square  root. 

Therefore,  if  a  number  has  no  divisor  equal  to  or  less  than  its 
square  root,  the  number  is  prime. 

542.  By  §  497,  IV,  the  number  of  combinations  of  n  different 

,,  .        .  1            i.     i.-        •    w(^  —  l)(w  —  2)  •••  to  r  factors 
things  taken  r  at  a  time  is  ^ ^-\ — ~-^ 

Since  the  number  of  combinations  of  n  different  things  taken  r 
itt  a  time  must  be  a  whole  number. 

Principle  3.  —  The  product  of  any  r  consecutive  integers  is  divisi- 
ble by  jr. 


THEORY  OF  NUMBERS  477 

n                           1     Tr-\i      w(w  —  l)(n  —  2) '••  (71  —  r  +  1) 
Corollaries.  —  1.  IJ  N  =  -^ ^ f ^^ ' — ^,  evei-y 

\r 

jn'inie  factor  of  the  numerator  that  is  greater  than  r  is  a  factor 

of  N. 

2.    If  n  is  a  positive  integral  exponent,  tlie  binomial  coefficients 

n   n(n  —  1)    n(n  —  1)  (n  —  2)  .  , 

-,  -!^ ^,  — ^^ '-^ ^,  •••,  are  integers. 

1'1.2'  1.2.3''  ^ 

l^XAMPLES 

1.  Show  that  ji^  —  w  is  an  even  number  when  n  is  integral. 

Proof 

The  integer  n  must  be  even  or  else  odd. 

Every  even  number  has  the  form  2p,  p  being  an  integer,  and  every  odd 
number  has  the  form  2 p  +  1  or  the  equivalent  form  2(p  —  1)+  1,  or2p  —  1. 

If  n  is  even,  let  n  =  2  p. 

Then,  n^  —  n  =  n(n  —  1)  =  2p(2p  —  1),  an  even  number. 

If  n  is  odd,  let  7i  =  2p  +  1. 

Then,  n^  —  n  =  n(n  —  1)  =  (2p  +  l){2p),  an  even  number. 

Hence,  whether  n  is  even  or  odd,  n^  —  n  is  even. 

Or,  again,  by  Prin.  3  the  product  of  any  two  consecutive  integers  is 
divisible  by  [2.     Hence,  n^  —  n,  or  n(n  —  1),  is  even. 

2.  Show  that  every  perfect  cube  belongs  to  one  of  the  forms 

7n,7n-\-l,7n  —  l. 

Proof 

Every  integer  must  belong  to  one  of  the  forms  7 p,  7 p  ±  I,  7 p  ±2, 
7p  ±3  ;  for  7p  +  4  =  7(p  +1)  —  3,  which  has  the  same  form  as  7  p  —  S, 
and  7p  —  4:  =  7(p  —  1)  +3,  which  has  the  same  form  as  7p  +  3 ;  and  simi- 
larly, 7  p  ±  5  and  7p  ±()  may  be  reduced  to  the  forms  7  p  ±2,  and  7  p  ±  1, 

respectively. 

(7p)8  =  348  p8  =  7(49p3):=  7  „. 

(7/)±l)3  =  343p3  J.  U7p^  +  2lp±  1 

=  7(49 p3  ±  21  p2  +  3 J,)  J.  1  =,  7  ,j  ±  1, 

(7j9  ±  2)3  =  343 1)3  ±  294  p2  +  84i9  ±  8 

=  7(49 j93  J.  42p^  +  \2p±  1)±  1  =  7n±  1. 

(7p  ±  3)3  =  343 p3  ±  441  p2  +  is9p  ±  27 

=  7(49p8±63p2  +  27p±4)T  1  =  7  n  T  1. 

Caution.  — Not  every  number  belonging  to  one  of  these  forms  is  a  perfect 
cube.    For  example,  7(6)  +  1,  or  43,  is  not  a  perfect  cube. 


478  ''W-;^va^^^  \^^F  NUMBERS 

3.   Find  what  rau  ,3  of  x  will  make  a^  —  3x  +  5  a 

perfect  square. 

Solution 

Let  x^-3x  +  b  =(.x  -  7n)2. 

Solving,  *  =  f*'-=^. 

2  n?.  —  3 

which  is  rational  for  all  rational  values  of  m. 

Thus,  substituting  •••,  -  4,  -  3,  -  2,  -  1,  0,  1,  2,  3,  4,  ...  for  m, 

x=...,  -1,  -1,1,1,1,4,  -  1,  I,  Jgi, .... 
Any  of  these  values  substituted  renders  x^  —  3  a;  +  6  a  perfect  square. 

4.  Find  which  of  the  numbers  431,  211,  253,  301,  623,  323  are 
prime  and  factor  the  others. 

5.  Of  the  first  ten  numbers  beginning  with  1001,  which  are 
composite  ? 

Prove  the  following : 

6.  The  sum  of  any  number  of  even  numbers  is  even. 

7.  The  sum  or  difference  of  two  odd  numbers  is  even. 

8.  The  difference  between  the  squares  of  two  odd  numbers  is 
divisible  by  8. 

9.  The  difference  between  a  number  and  its  cube  is  divisible 
by  6. 

10.  The  number  w'  —  5  n^  -f-  4  n  is  divisible  by  120. 

11.  Every  perfect  square  has  one  of  the  forms  3n,3n  +  l. 

12.  Every  perfect  square  has  one  of  the  forms  4  n,  4  n  +  1. 

13.  Every  prime  number  except  2  and  3  belongs  to  one  of  the 
forms  6  ?i  +  1)  6  n  —  1. 

14.  What  rational  values  of  x  will  make  a^  +  5  a;  +  2  a  perfect 
square  ? 

15.  What  values  of  x  will  rationalize  -\/ax  +  &  ? 


Suggestion.  —  Put  y/ax  +  b  =  m. 

16.    What  is  the  least  multiplier  that  will  make  33957  a  perfect 
square  ?  a  perfect  cube  ? 

Suggestion.  — Put  the  number  into  the  form  aPb^(f  •••  and  examine  the 
exponents  of  each  factor. 


(1) 


DETERMIN  rp' 

^king  constitueD*-"  *^^'^  xaiau\&T  l-.uo^* 


DETERMINANTS 


543.   Solving  the  simultaneous  independent  equations 

we  have  x  =  -^^ ^-i,     y  —  -^-^ ^-i* 

aib.2  —  a>J)i  a-Pi  —  ci-P\ 

Comparing  the  values  of  x  and  y  it  is  observed  that : 

1.  They  have  the  same  denominator. 

2.  The  numerator  of  the  value  of  x  may  be  formed  from  the 
denominator  by  replacing  the  coefficients  of  x  by  the  correspond- 
ing known  terms  k^  and  k^. 

3.  The  numerator  of  the  value  of  y  may  be  formed  from  the 
denominator  by  replacing  the  coefficients  of  y  by  the  correspond- 
ing known  terms  k^  and  k^. 

The  common  denominator  Uih^  —  api  is  called  the  determinant 
of  the  system. 

A  convenient  symbol  for  Oiftg  —  «2&i>  suggested  by  the  arrange- 
ment in  (1)  of  the  coefficients  of  x  and  y  in  two  columns  and  two 
rows,  is 

called  a  determinant  of  the  second  order.  ^ 

ai&2  —  ^2^1  is  called  the  developed  form,  or  the  development,  of 
this  determinant. 

0162  ^'Hd  —  a^i  are  called  its  constituents. 

Oj,  a2,  61,  &2  are  called  its  elements. 

Note.  —  Some  authors  employ  the  terms  eZemen^  and  constituent  with  the 
meanings  here.given  to  constituent  and  element,  respectively. 

479 


480  TTIEORY  6f  numbers 

544.  To  deveiop  5    •  ^^.Wne'seL'^^^  make  a^  — Sa- 

By  definition,  I   "*     '^  \=  a^b^  ■— a^b^- 

\(h     fh\ 

The  second  member  may  be  written  h-2ai  —  bia^i  or  —  bia2  +  Oib^,  etc. 

1.  The  positive  term,  afii  or  fta^D  is  obtained  by  multiplying  the 
element  aj  in  the  Jirst  column  and  first  row  by  the  element  62  ii^ 
the  next  column  and  next  row ;  or  by  multiplying  the  element  63  in 
the  second  column  and  second  row  by  the  element  ay  in  the  preced- 
ing column  and  preceding  row. 

The  selection  of  an  element  from  any  column  or  row  before  the 
selection  of  an  element  from  a  preceding  column  or  row  consti- 
tutes an  inversion. 

Then,  the  positive  term  formed  in  the  first  way  presents  no 
inversions,  but  formed  in  the  second  way  presents  two  inversions, 
namely,  the  selection  of  an  element  from  the  second  column 
before  that  of  an  element  from  the  preceding  column,  and  the 
selection  of  an  element  from  the  second  row  before  that  of  an 
element  from  the  preceding  row. 

In  either  case  the  positive  term  of  the  development  presents  an  even 
number  of  inversions. 

2.  The  negative  term,  —  a2&i,  or  —  ftia,?  is  obtained  by  multi- 
plying the  element  a.^  in  the  first  column  and  second  row  by  the 
element  b^  in  the  second  column  and  first  row,  and  making  the 
product  negative ;  or  by  selecting  the  elements  in  the  reverse 
order  and  making  the  product  negative.  In  the  first  way  there  is 
an  inversion  of  rows,  in  the  second  way,  an  inversion  of  columns. 

In  either  case  the  negative  term  of  the  developinent  presents  an  odd 
number  of  inversions. 

545.  Any  square  array  of  n^  elements  arranged  in  n  columns 
and  n  rows  represents  a  determinant  of  the  nth  order. 

In  harmony  with  the  principles  of  the  preceding  article  a  deter- 
minant of  any  order  is  now  defined  as  a  square  array  of  numbers 
that,  by  common  agreement,  represents  the  algebraic  sum  of  all 
the  products,  or  constituents,  that  can  be  formed  by  taking  one 
element,  but  not  more  than  one,  from  each  column  and  from  each 


DETERMINANTS 


481 


row,  making  constituents  that  present  an  eveii  number  of  inver- 
sions positive  and  constituents  that  present  an  odd  number  of 
inversions  negative. 

546.    Development  of  any  determinant. 

I  f\      ^1      Cl 

Let    a^    h-2    c.2   be  a  determinant  of  the  third  order. 

By  the  defiiaition  of  a  determinant,  each  constituent  of  this 
determinant  contains  three  elements  as  factors,  one  and  only  one 
taken  from  each  column  and  from  each  row. 

Hence,  the  constituents  involving  ai  are  Uih/ig  and  —  aib^Ci,  the 
latter  being  negative  because  it  presents  one  inversion.  There- 
fore, the  sum  of  the  constituents  involving  a^  is 

ai02Cs  —  ftiOgCo,  or  aj 

'-^3     ^3 

which  may  be  obtained  from  the  given  determinant  by  canceling 

or  deleting  the  elements  that  cannot  be  associated  with  %, 

Oi     -&J — Ci 

thus :  (jf2    &2    Ca 

.  (tg  O3  C3 


The  determinant  of  the  next  lower  order 


by  which  aj  is 


—  ^2 


multiplied  is  called  the  minor  of  the  element  a^.  When  the  minor 
is  given  the  proper  sign,  in  this  case  + ,  it  is  called  the  co-factor 
of  the  element. 

Similarly,  the  sum  of  the  constituents  involving  a^  is 

61      Ci 
derived  by  deleting  the  elements  that  cannot  be  associated  with  a^, 

thus :  «2    ^  ^ 

<p3    ^3    ^3 
and  giving  o^  the  sign  —  ,  because  in  each  constituent  ag  is  chosen 
before  an  element  of  the  preceding  row. 


In  tM..  case,  since 

02  is  D^^^'■^l^.'t■. 


=  a2X  — 


,  the  co-factor  of 


31 


482 


DETERMINANTS 


Similarly,  the  sum  of  the  constituents  involving  a^  is 

b.2    C2 

Since  each  constituent  of  the  given  determinant  must  involve 
either  a^,  a.^,  or  a^,  we  have  found  all  the  constituents.     Hence, 

«i 

The  same  result  is  obtained  by  nsing  any  column  or  any  row 
of  elements  as  the  first  column  is  used  above. 

For  example,  selecting  the  elements  of  the  second  column. 


61     Ci 

b2 

Ci 

&i 

Cl 

h 

Ci 

62    ~£s 

=  ai 

-  ttz 

+  013 

\0s 

C3 

&3 

C3 

h 

C2 

63     C3 

Oj 

6x 

Cl 

a2 

C2 

ttl 

Cl 

,        «! 

Cl 

«2 

62 

C2 

=  -6, 

+  62 

-&3 

ttg 

C3 

ttg 

C3 

^2 

C2 

«3 

&3 

C3 

=  —  ttoftiCg + agftjCa + aiftgCs — a3&2Ci — ai&3C2 + a A^d 
which  is  the  former  result  differently  arranged. 

The  above  discussion  applies  to  a  determinant  of  any  order. 
Hence, 

The  development  of  a  determinant  of  any  order  is  equal  to  the 
algebraic  sum  of  the  products  of  the  elements  of  any  cohimn  or 
row  and  their  respective  co-factors. 

547.  The  minors  corresponding  to  the  elements  aj,  Oa^,  •••, 
61,  &2)  •••)  are  denoted  by  A^,  A^,  •••,  B^,  B^,  •••. 

548.  Number  of  constituents. 

Since  the  co-factors  of  each  of  the  n  elements  in  any  selected 
column  or  row  of  a  determinant  of  the  ?ith  order  are  determinants 
of  the  (w  —  l)th  order,  a  determinant  of  the  nth  order  has  n  times 
as  many  constituents  as  a  determinant  of  the  (n  —  l)th  order; 
this,  in  turn,  has  {n  —  1)  times  as  many  constituents  as  a  deter- 
minant of  the  (n  —  2)th  order ;  and  so  on,  until  a  determinant  of 
the  2d  order  is  reached,  which  has  2  constituents.     Hence, 

A  determinant  of  the  yith  order  has  n(n  —  l)(n  —  2)  •••  2,  or  [w, 
constituents. 


DE  TERMINA  NTS 


483 


1.   Develop  the  determinant 


Examples 

6      9  8 

10    11  12 

14    15  16 

Solution.  —  Multiplying    the    elements    of    the    first    column  by  their 
co-factors,  and  adding,  the  given  determinant  is  reduced  to 

111191  IQ        9t  \  IQ        9i\ 

15   lel-^His   lel+i^lii    i2h-^-2^«  +  280  =  i6. 

12  3  4 
5  6  9  8 
9  10  11  12 
3  14  15  16 
Solution.  — Proceeding  as  in  Ex.  1,  the  given  determinant  is  reduced  to 


2.   Develop  the  determinant 


6   9 

8 

10  11 

12 

-5 

14  15 

16 

2   3 

4 

10  11 

12 

+  9 

14  15 

16 

2 

3 

4 

6 

9 

8 

-3 

4 

15 

16 

10     11     12 


Since  by  Ex.  1  the  first  determinant  is  equal  to  16,  the  given  determinant 
=  16-5. 2I!?    !5|-5(-10)L!    .!l-5.14l    ^      41 


15     16 
9      8 


15     16 


+  ^'2li5   lel  +^(-«)|i5  lel  +  ^-i^ 


11   12 

3     41 


3.2 


,..41  134 

.Sf—e^  —3-10 

^     ^    11     12        ^^"98 


11     12 

=  16  +  40  -  600  +  560  +  432  +  648  -  1512  -  120  -  144  +  360 
=  -  320. 


Develop  the  following  determinants : 


4 

9 

2 

3 

5 

7 

8 

1 

6 

1 

7 

1 

3 

3 

3 

5 

1 

5 

3 

2 

0 

0 

6 

4 

1 

1 

1 

2 

2 

3 

4 

3 

2 

2 

5. 


3  2    1 

4  8    3 

5  2    1 


3  4    2    5 

0  3    12 

0  12    1 

2  0    2    7 


1111 

0    4    2  7 

0    3    2  7 

0    2    2  1 

2  0    0  0 

3  2  2  1 
2  12  1 
13    5  1 


*  For  economy  of  space  the  sign  of  a  negative  element  may  be  written 
above  the  element. 


484 


DETERMINANTS 


^ 


10.  Express  ain? —  2  a  —  hmn -\-2hc-\-mx  —  ncx  as  a  determi- 
nant. 

Solution 

Since  there  are  6,  or  [3,  constituents,  it  is  likely  that  the  required  determi- 
nant is  of  the  third  order,  and  that  the  terms  —  2  a  and  mx  have  a  factor  1 
or  —  1  unexpressed. 

avP'  —  2a  —  bmn  +  2bc  +  mx  —  ncx 

=  a(n  •  H  —  2  •  1)—  6(m  •  n  —  2  •  c)+  x(m  •  1  —  n  •  c) 


(a2_62)_c2. 
1-  (x^  +  1). 
m^  —  (n^  —  n). 


Express  as  determinants ; 


a 

m 

c 

n 

1 

m 

c 

m 

c 

-b 

+  X 

^ 

b 

n 

1 

2 

n 

2 

n 

n 

1 

X 

2 

n 

11. 

12. 
13. 
20. 
21. 
22. 


14. 
15. 
16. 


a  +  X. 
b'  +  l. 


25  -  21. 

42  +  33. 

ab  —  cd. 

^liVi  -  Vs)  +  XiiVs  -  Vi)  +  x^iVx  -  Ih)- 
a^  —  abc  —  abc  +  6^  -f  c^x  —  abx. 
abc  —  axy  —  acx  -f-  xyz  -\-  abx  —  bh. 


17. 
18. 
19. 


23.  a 


3  a 

4  2 

5  a 


+  c 


REDUCTION  OF  DETERMINANTS 
549.   A  determinant  that  is  equal  to  zero  is  called  a  vanishing 
determinant. 


550.    1.   How  does 


compare  in  form  and  value  with 


5    2 
4    3 

9 

7 
2 

4 
1 

with 

L 

7     2 
4     1 

9 

6 
8 

9 
4 

w 

:th 

6    8 
9    4 

2    8    1 

2    3    7 

Ol 

6l 

Ci 

tti     ttg     ttj 

3    5    6 

with 

8    5    3 

9 

tta 

b. 

C2 

with 

bi     &2     6j 

7    3 

4 

1 

6    4 

ttg 

&3 

Cs 

Ci 

Cg          Cj 

2.  How  is  the  value  of  a  determinant  affected  by  changing  the 
rows  into  columns  and  the  columns  into  rows  ? 

Principle  1.  —  Tlie  value  of  a  determinant  is  not  changed  by 
changing  its  columns  into  rows  and  its  rows  into  columns,  provided 
that  their  order  of  succession  is  not  changed. 


DETERMINANTS 


485 


The  above  principle  may  be  established  as  follows : 

Since  the  1st,  2(i,  •••,  nth  columns  become  the  1st,  2d,  •••,  wth  rows,  re- 
spectively, and  vice  versa,  the  relative  position  of  the  elements  is  not  changed. 

Therefore,  each  element  of  any  column  or  row  has  the  same  co-factor  as 
before  the  reduction.     Hence,  the  value  of  the  determinant  is  not  changed. 

Corollary.  —  Wluitever  is  true  of  the  columns  of  a  determinant 
is  true  of  its  rows,  and  vice  versa. 


551. 


1.    What  is  the  value  of 
0    3 
^     of 


of 


3 

2 

5 

0 

0 

0 

4 

1 

6 

2.  What  is  the  value  of  a  determinant  if  all  the  elements  of 
one  column  or  row  are  zeros  ? 

Principle  2. — A  determinant  that  has  one  or  more  columns  or 
rows  of  zeros  is  equal  to  zero. 

For  since  each  constituent  must  have  for  a  factor  an  element  of  the 
column  or  row  whose  elements  are  zeros,  each  constituent  is  equal  to  zero. 


552.   1. 

9     4  I 

15  8  r 


How  does 
with 


12 


compare  in  form  and  value  with 
3 


with 


2.  What  is  the  effect  of  multiplying  or  dividing  all  the  elements 
in  a  column  or  row  by  the  same  number  ? 

Prixciple  3.  —  Multiplying  or  dividing  all  the  elements  in  a 
column  or  row  of  a  determinant  by  the  same  number  multiplies  or 
divides  the  determinant  by  that  number.    (§  85,  §  104,  3.) 

Corollary.  —  Changing  the  signs  of  all  the  elements  in  one 
column  or  row  changes  the  sign  of  the  determinant. 


553.    1.    How  are 


and 


formed  from 


How  do  they  compare  with  the  latter  in  value  ? 


2.    Show  that 


3 

7 
10 


3 

7 
10 


3 

7 
10 


Principle  4.  —  The  interchange  of  any  two  columns  or  of  any 
two  rows  of  a  determinant  changes  the  sign  of  the  determinant. 


486 


DETERMINANTS 


The  above  principle  may  be  established  as  follows : 

Let  Z)  be  a  determinant  of  the  nth  order  and  D'  a  determinant  formed  by 
interchanging  any  two  columns  of  D. 

It  is  to  be  proved  tliat  D'  =—  D. 

By  the  definition  of  a  determinant,  §  545,  the  elements  forming  each  con- 
stituent may  be  selected  from  the  columns  in  any  order  we  please,  taking  one 
but  not  more  than  one  fi-om  each  column  and  row,  provided  each  constituent 
so  formed  is  given  the  proper  sign  showing  the  even  or  odd  number  of  inver- 
sions of  the  established  order  of  columns  and  rows. 

Then,  let  the  last  two  elements  of  each  constituent  be  chosen  from  the  two 
columns  to  be  interchanged  in  the  order  in  which  these  columns  stand,  giv- 
ing the  result  the  proper  sign.  By  this  method,  when  the  columns  have  been 
interchanged,  each  constituent  will  have  one  more  inversion  than  before, 
namely,  the  inversion  in  the  order  of  the  last  two  columns. 

Hence,  the  sign  of  each  constituent  will  be  changed  by  interchanging  the 
two  columns,  and  by  the  Distributive  Law  this  changes  the  sign  of  D ;  that 
is,  D'  =-D. 

554.  By  changing  places  successively  with  each  of  the  preced- 
ing columns,  any  column  may  be  made  the  leading  column,  pro- 
vided, Priu.  4,  that  when  the  number  of  columns  supplanted  by  the 
advancing  column  is  odd  the  sign  of  the  determinant  is  changed. 

Since  the  same  is  true  of  the  advance  of  any  row  to  the  position 
of  leading  row,  any  element  may  be  brought  to  the  position  of 
leading  element  by  a  proper  number  of  advances  of  its  column  and 
row,  provided  that  the  sign  of  the  determinant  is  changed  when 
the  sum  of  the  number  of  columns  and  the  number  of  rows  preced- 
ing the  column  and  row  in  which  the  element  stands  is  odd. 

Therefore,  since  the  co-factor  of  the  leading  element  is  always 
positive,  the  sign  of  the  co-factor  of  any  element  is  -\-  when  the  com- 
bined number  of  cohtmns  and  rows  preceding  the  column  and  row  of 
the  element  is  even,  and  —  when  this  number  is  odd. 

12      3 
Thus,  in  the  determinant   4      5 
7      8 
negative  ;  of  5,  positive  ;  of  6,  negative  ;  of  7,  positive  ;  etc. 

555.  The  preceding  principle  suggests  a  device  for  developing 
a  determinant  of  the  third  order. 


the  co-factor  of  4  is  negative  ;  of  2, 


In 


ttj     bi 
a^    &2 


draw  diagonals,  thus 


DETERMINANTS 


487 


The  constituents  ai^gCs  ^-nd  —  agftoCi,  whose  elements  lie  on  the 
diagonals,  are  called  the  principal  diagonal  and  the  secondary  diago- 
nal, respectively.  In  a  determinant  of  the  third  order  the  principal 
diagonal  is  posiYive  and  the  secondary  diagonal  is  negative. 

To  find  the  other  positive  and  negative  constituents,  by  two 
interchanges  of  columns,  and  again  by  two,  we  have 


Oi 

61 

Ci 

h 

Ci 

Oi 

Ci 

«! 

61 

a^ 

h 

C2 

= 

h 

C2 

^2 

= 

C2 

02 

&2 

ttg 

h 

C3 

h 

C3 

ag 

C3 

ttg 

63 

The  principal  diagonals  of  these  determinants  are  the  three 
positive  constituents  of  the  given  determinant  and  the  secondary 
diagonals  are  the  three  negative  constituents. 

The  three  equal  determinants  and  their  diagonals  are  written  i-n 
the  form  + 


in  which  the  principal  diagonals  are  positive  and  the  secondary  di- 
agonals are  negative. 

Caution.  —  This  device  does  not  apply  to  determinants  of  a  higher  order 
than  the  third. 


556.   1.  What  is  the  value  of 

7 
of 


?  of 


10 
6 
4 


2.  Form  other  determinants  each  with  two  columns  or  rows 
alike  or  differing  by  a  constant  multiplier.    What  value  has  each,? 

Principle  5.  —  If  the  corresponding  elements  in  any  tivo  columns 
or  rows  of  a  determinant  are  the  same,  or  if  the  elements  in  one  col- 
umn or  row  are  equimultiples  of  the  corresponding  elements  in  the 
other,  the  determinant  is  equal  to  zero. 

The  above  principle  may  be  established  as  follows ; 

1.   Let  D  be  a  determinant  having  two  identical  columns  or  rows. 

By  Prin.  4,  if  these  two  cohimns  or  rows  are  interchanged  the  sign  of  the 
determinant  will  be  changed,  giving  —D.  But  since  the  two  columns  or 
rows  are  identical,  interchanging  them  does  not  change  the  determinant. 

Hence,  D  =  -  D.     But  D  =  -  D  only  wlien  Z)  =  0.     Therefore,  Z)  =  0. 


488 


DE  TERM  IN  A  NTS 


2.  Let  the  elements  in  one  column  or  row  be  m  times  the  corresponding 
elements  in  another  column  or  row,  and,  Prin.  3,  let  the  determinant  be 
represented  by  mD. 

Then,  as  in  1,  mD  =  —  niD, 

which  is  true  only  when  Z)  =  0,  for  m  is  not  equal  to  zero. 

557.    To  what  determinant  of  the  second  order  is 


ai 

6i 

Ci 

0^2 

&2 

C2 

as 

bs 

Cs 

eqital  if  ag  =  0  and  ag  =  0  ?  if  Z>i  =  0  and  Cj  =  0  ?  if  &2  =  f>  and 
Co  —  O?  if  all  the  elements  but  one  in  any  column  or  row  are 
equal  to  zero  ? 

Principle  6.  —  If  all  the  elements  but  one  in  any  column  or  row 
of  a  determinant  are  equal  to  zero,  the  determinant  is  equxil  to  a 
single  determinant  of  the  next  lower  order,  namely,  the  product  of 
the  element  and  its  co-factor. 

For  each  of  the  co-factors  corresponding  to  the  other  elements  in  that 
column  or  row  has  the  coefficient  0,  and  so  becomes  0. 

558.  By  Prin.  6,  any  determinant  may  be  written  as  the  minor 
of  the  element  1  or  —  1  of  a  determinant  of  the  next  higher  order 
equal  to  the  given  determinant,  provided  that  the  other  elements 
in  the  same  column  or  row  as  1  or  —  1  are  zeros. 


Thus, 


b 
d  ~ 

1     *    * 
0    a    6 

1   or 

a    b    * 
c     d    * 

,    or 

*    a    b 
i    0    0 

0    c    d 

0     0     1 

*    c    d 

m  which  each  asterisk  stands  for  any  finite  number. 


559.   1.   IfZ>  = 


,  showthati)=(5a-3c)  +  (56-3d). 
3  d  as  determinants,  each  having 


a  -\-h     3 
c  +  d    5 

2.  Write  5  a  —  3  c  and  5  h  ■ 
the  same  second  column  as  2). 

3.  Into  what  two  determinants,  then,  may  D  be  resolved  ? 

Principle  7.  —  If  each  element  of  any  column  or  row  of  a  deter- 
minant is  compound,  the  determinant  may  he  written  as  the  algebraic 
sum  of  two  or  more  determinants.     (§  85.) 

560.  It  follows  from  Prin.  7  that  if  two  or  more  determinants 
differ  only  in  the  elements  of  one  column  or  row,  they  may  be 
united  into  a  single  determinant. 


DE  TERMINANTS 


^89 


Thus, 


5     2     3 

-3    2    3 

5-3    2     3 

2     2     3 

2    4     3 

+ 

1    4    3 

= 

2  +  143 

= 

3    4     3 

1     5     4 

-15    4 

1-1     5     4 

0     5    4 

561.   1.   Separate 


2.    Separate 


5-3    3    10 
6-4    4    20 

8-7     7    30 
is  the  value  of  the  second  determinant  ? 

5-3     3 
6-4    4 

-7     7 


10- 

-12 

20- 

-16 

30- 

-28 

into  two  determinants.    What 


into  four  determinants.   What 


5 

3     10 

6 

4    20 

8 

7     30 

is  the  value  of  each  ?     Then,  what  simpler  form  has 


Principle  8.  —  If  the  elements  of  any  column  of  a  determinant 
are  increased  or  diminished  by  the  corresponding  elements  or  by 
equimultiples  of  the  corresp)onding  elements  of  any  other  column,  the 
value  of  the  determinant  is  not  changed. 

Tlie  same  is  true  of  any  ttvo  rows. 

The  above  principle  may  be  established  as  follows : 

ffli    6i    •••    ki 
02    b-z    •••    k^ 


Let 


bn 


kfi 


be  any  determinant, 


and  let  m  be  any  positive  or  negative  number. 

1.    Suppose  that  the  elements  of  the  second  column  are  multiplied  by  m 
and  added  to  the  corresponding  elements  of  the  first  column. 
Then,  by  Prin.  7,  the  resulting  determinant  is  resolved  thus  : 


Oi  4-  mbi    bi     • 
0(2  +  fnbfi    bz     • 

.     ki 

•     kz 

= 

«!     bi     •••     ki 

az    bz    •■'     kz 

an      bn      •••      *„ 

D  +  0  =  D. 

+ 

mbi     bi     • 
mbz     bz     • 

mbn    bn    ■ 

■  ki 

■  kz 

a„  +  mbn    bn    • 

.5, 

••      kn 

'  •      ^n 

Prin 

2.  Let  the  modified  column  be  any  column  after  the  first,  say  the  ?*th. 
Then,  by  r  interchanges  of  columns  the  modified  column  may  be  made 

the  leading  column,  and  the  determinant  may  be  resolved  as  in  1,  into  B  +  O 
or  —  D  +  0,  according  as  the  number  of  columns  preceding  the  rth  is  even 
or  odd.  In  either  case  by  restoring  the  leading  column  to  its  original  posi- 
tion the  result  obtained  will  be  Z),  the  given  determinant. 

3.  A  similar  proof  may  be  given  for  modifying  any  row. 


490 


DETERMINANTS 


1.    Evaluate  the  determinant 


Examples 

2    3     4       1 

4  2     12 
112      3 

5  0    3    10 


Solution 


2 

3 

4    1 

4 

1 

2 

1 

1  2 

2  8 

= 

5 

0 

3  10 

1     0     2  10 

6    0    3  8 

112  3 

5    0     3  10 


# 

1     2     10 

1     2     IT) 

_ 

6     3      8 

_ 

0    9     52 



9     62 

5     3     10 

0     7     40 

7     40 

=  4. 


Explanation.  —  The  aim  is  to  reduce  each  determinant  in  turn  to  a  de- 
terminant of  the  next  lower  order  (Prin.  6)  by  adding  such  multiples  of  the 
elements  of  some  column  or  row  to  the  corresponding  elements  of  one  or 
more  other  columns  or  rows  (Prin.  8)  that  all  of  the  elements  but  one  in  some 
row  or  column  of  the  resulting  determinant  shall  be  zeros.  The  column  or 
row,  multiples  of  whose  elements  are  added  (or  subtracted),  may  be  called 
an  operating  column  or  row  and  is  marked  with  an  asterisk. 

Thus,  selecting  the  third  row  for  an  operator,  we  subtract  3  times  the 
operator  from  the  first  row,  obtaining  T  0  2  10  ;  and  add  2  times  the  operator 
to  the  second  row,  obtaining  6  0  3  8.  The  operator  itself  must  be  btought 
down  unchanged,  in  order  that  the  parts  added  or  subtracted  may  be  vanish- 
ing determinants. 

Since  all  the  elements  except  —1  in  the  second  column  of  the  resulting 
determinant  are  zeros,  this  determinant  (Prin.  6)  is  equal  to  —1  times  its  co- 
factor,  which  is  negative,  because,  §  554,  the  element  —  1  is  preceded  by  ele- 
ments in  an  odd  number  of  columns  and  rows.  Hence,  —  1  times  this  negative 
co-factor  gives  a  positive  determinant  of  the  third  order. 

Continuing  this  process,  the  result  obtained  is  4. 


2.    Show  that 


Prin.  8  and  5, 


14     7 

2  5    8 

3  6    9 


is  a  vanishing  determinant. 
Solution 


1    4    7 

1     3    6 

2     5    8 

= 

2     3    6 

3     6    9 

3     3    6 

3.   Evaluate  the  determinant 


3 
5 

9     11 
5      3 


=  0. 

4 

7 


2     3 


5 

9 

13 

2 
6 


DETERMINANTS 


491 


Solution 


12  3  4  5 

2  4  5  7  9 

3  5  9  11  13 
2  4  5  3  2 
12  3  5  6 


0 
0 

0 

1 

0 

1 

0 

1 

1 

0 

1 

2 

= 

0 

1 

5 

8 

0 

0 

1 

1 

0     111 

1     1     1 

10     12 

_ 

1     5     8 

0     15     8 

0     1     1 

0    0     11 

* 

1     0    0 

4     7 

1     4     7 

=  — 

1     1 

0     1     1 

=  -3. 


Evaluate  the  following ; 


5. 


8 

4 

6 

2 

2 

4 

2 

3 

4 

4  2  12 

2  3  2  5 

3  2  12 

5  6  4  9 

5  2  7  5 

6  3  14 

4  2  13 
6  3  2  5 

4  12  1 

5  2  3  1 

2  112 

3  2  4  6 


10. 


2  4  4  6  1 

2  5*312 

3  112  1 
2  1112 
5  2  2  3  1 

2  12  3  3 
12  2  2  4 

3  2  13  2 

2  3  4  0  2 

3  3  2  3  0 


a  1  1  1 
1  a  1  1 
1  1  a  1 
Ilia 


562.    To  factor  a  determinant. 


1.    Factor  Z)  = 


Examples 

X  b  b 

X  a  c 

-y  a  c 


Solution.  —  If  a  —  c,  the  second  and  third  columns  are  identical  and, 
Prin.  5,  the  determinant  vanishes.  Hence,  by  the  Factor  Theorem,  §  136, 
a  —  c  is  a  factor  of  D. 


492 


DETERMINANTS 


Again,  \l  x=  —  y,  the  second  and  third  rows  are  identical  and  P  ^0. 
Hence,  x  +  y  is  a  factor  of  D. 

Since  every  constituent  of  D  is  of  the  third  degree  and  (a  —  c)  (x  +  y)  is 
of  the  second  degree,  D  must  have  another  factor  of  tlie  first  degree.     Substi- 

a 


tuting  0  for  6,  D  is  equal  to  x  times  the  co-factor 


,  VFhich  is  equal  to  0; 


that  is,  Z>  =  0.  Hence,  the  other  factor  of  Z)  is  6  —  0,  or  6  ;  or  it  may  be  —h. 
It  remains  to  find  wliether  D  —  b  {a  —  c)  {x  +  y)  or  —  6  (a  —  c)  (x  +  2/). 
The  secondary  diagonal  of  Z)  is  +  aby  and  this  is  the  only  constituent  of 
D  involving  a,  6,  and  y.  Since  the  sign  of  aby  is  +  in  ?>  (a  —  c)  (v  +  y) 
but  —  in  —  6  («  —  c)  (x  +  2/),  Z)  =  6  (a  —  c)  (x  +  y). 


Factor  the  following  determinants  by  inspection : 


X 

1 

h 

y 

1 

a 

X 

1 

a 

3. 


a^    a     1 

b^    b     1 
c2     c     1 


2    2    5 

2x5 
X    S    5 


563.    Solution  of  simultaneous  simple  equations. 

It  has  been  shown  that  in  a  system  of  two  simultaneous  simple 
equations  of  the  form  ax  -{-by  =  c,  either  unknown  number  is  equal 
to  a  fraction  whose  denominator  is  the  determinant  of  the  system  and 
whose  numerator  is  the  determinant  of  the  system  with  the  known 
terms  substituted  for  the  corresponding  coefficients  of  that  unknown 
number. 

By  trial,  the  principle  is  found  to  hold  for  the  solution  of  three 
simultaneous  simple  equations. 

j-  axx  +  byy  +  Ci^  =  Ai, 

Thus,  given  j  a^x  +  h^y  +  C'lZ  =  ki^ 

\  a?,x  +  bzy  +  csz  =  ^3. 

Solving  by  the  ordinary  process  of  elimination,  then  rearranging  and 
grouping  terms, 

^__  k-i  (&2C3  -  &3C2)  -  ki  (ftiC3  -  fesPi)  +  ^'3  JbxC-i  -  &2C1) 
«i  (62C3  —  63C2)  -  ai  {b\Gz  -  bsci)  +  as  (61C2  —  boCi) 


ki 

bi 

Cl 

ki 

&2 

C2 

ks 

63 

Cs 

«! 

61 

Cl 

az 

&2 

C2 

as 

63 

C3 

Similarly  for  the  values  of  y  and  ». 


DETERMINANTS 


493 


The  principle  will  now  be  proved  to  be  general ; 


Let 


aix  +  biy  +  ciz  +  •••  =  Ai, 
a2X  +  b-iV  +  c-iZ  +  ■■■  =k2, 


(1) 


i  a„x  +  ft»y  +  c„2  H =  An 

be  a  system  of  n  simple  equations.  Let  D  represent  the  determinant  of  the 
system,  Di  the  determinant  of  the  system  with  the  known  terms  fci,  k^^,  ••• 
substituted  for  the  corresponding  coefficients  of  x,  and  ^i,  A^^  •••  the  co-fac- 
tors of  «!,  02,  •••• 


Then, 


D 


ax 

hi 

Ci       ••• 

ki 

bi 

Ci     .. 

<H 

h» 

C2       ••• 

,  and  Dx  = 

k.2 

62 

C2       •• 

a„ 

bn 

C„      ••• 

Kn 

b„ 

c„    •• 

(2) 


Multiplying  the  first  equation  of  the  system  by  Ai,  the  second  by  A2,  etc., 
and  adding  the  resulting  equations, 

(aiAi  +  aoAz  +  — f-  a„^„)x 


=  kiAi  +  koA^  +  •••  +  k„A„ 


(3) 


+  ibiAi  +  62^2  + 1-  b„An)y 

+  ■  . 

Since  the  coefficient  of  x  in  (3)  is  the  sum  of  the  products  of  the  elements 
in  one  column  of  D  and  their  co-factors,  the  coefficient  of  x  is  equal  to  D, 
and  the  second  member  of  (3)  is  equal  to  Dx-  The  coefficient  of  y  in  (3) 
differs  from  that  of  x  only  in  having  the  elements  of  the  second  column  of  D 
repeated  in  the  first  column,  bi,  bz,  •■•  replacing  ai,  ao,  •••,  thus  : 

61  61      Ci 

62  &2       C2 


bn      b„      Cn      ■ 

By  Prin.  5,  this  determinant  is  equal  to  zero,  and  in  like  manner  the  coeffi- 
cients of  the  other  unknown  numbers  vanish.     Hence,  (3)  becomes 


Dx  =  Dx 


whence,  x  =  — 2- 
D 


Similarly,  Dy  =  Dy-,  whence,  y  = 

So  for  each  unknown  number. 

Examples 

Solve  the  following  by  determinants : 

2x-t5y  =  9, 
3x  +  2y  =  S. 

Sx  +  2y  =  12, 
4:X  +  3y  =  17. 


D 


(4) 


1. 


2  a;  +  7  ?/  =  30, 
a;  +  4  ^  =  17. 

\bx-\-    y  =  12, 


[2x-^Sy  =  10, 


494 


DE  TERMINA  NTS 


8. 


9. 


10. 


4a;-3?/  =  8, 
x  +  4y  =  21. 

3x-2y=-2, 
2x-3y=-5. 

( ax  +  by  =  c, 
[  mx  +  ny  =  d. 

ax  —  by=r, 
cx-\-  dy  =  s. 

2x  +  5y  +  2z  =  27, 
8x-\-6y  +  3z  =  A6, 
3x  +  7y  +  5z  =  47. 

9x  +  2y-\-z  =  25, 
5x-\-y-\-z  =  14:, 

7x  +  Sy  +  2z  =  25. 


(  (a  +  h)x  —(a  —  b)y  =  4:  ab, 
y(a-b)x+{a  +  b)y  =  2a?~2b\ 

{3x  +  2y  +  3z  =  17, 

12.  I  2a;  +  ?/  +  2  2  =  10, 
[5x-\-5y-{-z  =  29l. 

'  2  a;  +  3  2/  -  4  2  =  18, 

13.  •  x  +  y-\-z  =  12, 

.5x  —  y  —  z  =  12. 

(x  +  2y  +  z  =  0, 

14.  ^  2x  +  y  +  z  =  2a  —  b, 
x—y  —  2z  =  3b. 

x-{-y  =  2  a, 
lb.  \  y  +  z  =  3a  —  bf 
[z  +  x  =  3a. 


16. 


17. 


u  —  x  +  2y  —  3z=—5, 
3u~x-[-y  —  2z  =  2, 
2u  +  x-{-y  —  z  =  9, 

I  —5u  +  2x  —  7y  +  z=  —  12. 

'2u  +  3v  —  4:X  +  y  =  0, 

u  —  v-\-x  —  y=  —2, 

7  ti  +  2v  —  3x  +  y  =  %, 
I  5  t<  +  8  i;  - 10  a;  +  3  ?/  =  3. 

564.  An  equation  in  which  every  term  is  of  the  first  degree  in 
some  unknown  number  is  called  a  Homogeneous  Linear  Equation. 

ax  =  by.,  or  ax  —  by  =  0,  is  a  homogeneous  linear  equation. 

565.  By  §  563  the  denominator  of  the  value  of  each  unknown 
number  in 

[  a^x -{- b^y  +  c-^z -\ =  0, 

agaj  +  da^ +  C2Z+ •••  =0,  (1) 


.  a„a;  +  6„2/  +  c„2  + 


0 


I 


DETERMINANTS 


495 


is  the  determinant  of  the  system,  and  the  numerator  is  the  same 
determinant  with  0  substituted  for  each  cqelficient  of  the  unknown 
number.  Therefore,  the  numerator  in  each  case  will  have  one 
column  composed  entirely  of  zeros,  and,  Prin.  2,  will  be  equal  to  0. 

0  0  0      , 

.'.  0;  =  —,  y  =  — ,  z  =  — ,  etc. 
D   ^      D  D 

Hence,  each  unknown  number  is  equal  to  zero,  except  when 
i>  =  0,  in  which  case  each  is  indeterminate  and  the  system  is 
indeterminate. 

The  case  in  which  i)  =  0  is  the  case  in  which  the  equations  in 
(1)  are  not  independent.  For  if  it  is  possible  to  form  any  equa- 
tion in  (1)  by  combining  multiples  of  two  or  more  of  the  other 
equations  by  addition  or  subtraction,  it  is  possible  to  make  two 
rows  of  D  identical  by  the  same  process. 

Hence,  if  n  homogeneous  linear  equations  involving  n  unknown 
numbers  are  independent,  the  unknown  numbers  are  separately  equal 
to  zero. 


566.  A  system  of  w  —  1  independent  linear  equations  involving 
n  unknown  numbers  is  indeterminate  (§  214,  proof)  ;  but  if  the 
equations  are  homogeneous,  the  ratio  of  any  two  unknown  num- 
bers may  be  found. 


Thus,  let  a\X  -\-  h\y  -f  Ciz  =  0, 

and  a2X  +  h^y  -f  C2«  =  0 

be  given,  to  find  the  ratios  of  x  to  y,  x  to  0,  and  y  to  z. 

X  .  ^  y 


From  (1), 
From  (2), 

Solving, 
also,  from  (5), 


z         z 


02-4-62^  =  - 

z         z 


C2- 


-Cl 

61 

a\ 

-Cl 

-  C2 

K 

and   ^  = 

z 

ai 

-  C2 

«! 

61 

«i 

fti 

0,1 

62 

a2 

62 

-Cl       fti 

y 

—  C2     62 

z 

Ol      —  Cl 

ai     —  C2 

6i 

Cl 

62 

C2 

Cl 

ax 

C2 

a% 

(1) 

(2) 

(3) 
(4) 

(6) 


CONYEKGENCY   OF   SERIES 


567.    By  division —i— =  1  +  cc  +  0^2  _|_  ^  _l ^  (i^ 

an  infinite  series  the  sum  of  whose  first  n  terms  is 

l+x  +  x^  +  a^-\ H  ^•"  =  ^—^^ 

1  —X 

Therefore,  the  diiference  between  the  fraction  from  which  the 
series  arose  and  the  sum  of  the  first  n  terms  of  the  series  is 

1         1  — a;"  x" 

or 


1—x      1  —  x'       1  —  x 

As  n  increases  without  limit,  this  difference  decreases  numeri- 
cally without  limit  if  x  is  numerically  less  than  1,  but  increases 
numerically  without  limit  if  a;  =  1  or  is  numerically  gi-eater  than  1. 

Hence,  if  aj'is  numerically  less  than  l,sthe  sum  of  the  first  n 
terms  of  the  series  approaches  the  value  of  the  fraction  as  a  limit, 
but  if  ic  =  1  or  is  numerically  greater  than  1,  the  sum  of  the  first 
n  terms  diverges  farther  and  farther  from  the  value  of  the  fraction 
as  n  increases,  and  approaches  no  fixed  value  as  a  limit. 

li  x=  —\,  the  series  becomes 

1-1  +  1-1  +  -, 

the  sum  of  whose  first  n  terms  oscillates  between  1  and  0  accord- 
ing as  n  is  odd  or  even,  while  the  value  of  the  fraction  is  ^. 
It  is  evident  from  the  above  discussion  that  the  series 

1+X  +  X^  +  'J^+--- 

stands  for  no  definite  fixed  number  unless  x  is  numerically  less 
than  1,  in  which  case  the  sum  of  the  first  n  terms,  as  n  is  indefi- 
nitely increased,  converges  toward  the  fixed  number 

J-  —  X 
496 


CONVERGENCY  OF  SERIES  497 

568.  When  the  sum  of  the  first  n  terms  of  an  infinite  series 
approaches  a  constant  number  as  a  limit,  as  n  is  indefinitely 
increased,  the  limit  is  called  the  sum  of  the  series,  and  the  series 
is  called  a  Convergent  Series. 

As  n  increases  without  limit,  the  sum  of  the  first  n  terms  of  the  series 
l  +  |  +  ^  +  ^  +  --.  approaches  2  as  a  limit.  Hence,  2  is  the  sum  of  the  series, 
and  the  series  is  convergent. 

The  series  1  +  a;  +  x^  +  x^  +  •••  is  convergent  for  all  values  of  x  numeri- 
cally less  than  1. 

569.  A  series  that  is  convergent  without  regard  to  the  signs  of 
its  terms  is  called  an  Absolutely  Convergent  Series. 

The  series  \-\  +  \-\j^  ...,  i  +  i  +  |+i  +  ...,  _i_i_i_i 

are  absolutely  convergent ;  for  even  when  the  signs  of  all  the  terms  are  alike, 
the  sum  of  the  first  n  terms  remains  finite  however  great  n  becomes.  But 
the  series  '^  —  \  +  \  —  \  +  •"■,  though  convergent,  is  not  absolutely  convergent, 
since  l  +  ^  +  ^  +  l^  +  .-is  not  convergent,  as  will  be  shown  later. 

570.  When  the  sum  of  the  first  n  terms  of  an  infinite  series 
can  be  made  numerically  greater  than  any  finite  number  by  tak- 
ing n  sufiiciently  great,  the  series  is  called  a  Divergent  Series. 

The  series  1  +  a;  +  a:^  4.  x^  -f  ...  is  divergent  if  x  =  1  or  any  number  numeri- 
cally greater  than  1,  as  2  or  —2  or  3,  etc. 

571.  When  the  sum  of  the  first  n  terms  of  an  infinite  series 
oscillates  between  certain  fixed  values,  as  n  is  indefinitely  in- 
creased, the  series  is  called  an  Oscillating  Series. 

The  series  \  +  x  -\-  x-  -{-  x^  +  •■•  is  oscillating  if  x  =  —  1,  for  the  sum  of  the 
first  n  terms  oscillates  between  1  and  0  according  as  n  is  odd  or  even. 

The  series  l-l-2-l  +  l-l-2-f-l-l-2-l-|-l-|-2-f-..  is  oscillat- 
ing, for  the  sum  of  the  first  n  terms,  as  n  takes  the  values  1,  2,  3,  •••,  is  one 
of  six  fixed  values  that  recur  in  the  order  1,  0,  —2,  —3,  —2,  0. 

572.  Any  infinite  series  may  be  represented  by 

Wi  +  W2  +  «3  H \-u^-\ , 

in  which  «i  denotes  the  first  term,  Wg  the  second  term,  etc.,  and  w„ 
the  nt\v  term,  or  any  term.  Since  n  may  be  given  the  successive 
values  1,  2,  3,  •••,  a  series  may  be  represented  by  its  nth  term. 

ADV.  ALG. 32 


498  CONVERGENCY  OF  SERIES 

Theseries  ? +  ^  +  ^  +  ^ +...+  2?-±I-f  ...  is  represented  by  ^^^^:    the 
1      2     3     4  71  n 

series  1-1  + 1- 1+ ...  +  (_i)n-ii  +  ...  is  represented  by  (-l)»-il. 
1234  n  n 

The  sum  of  the  first  n  terms  of  a  series  is  represented  by  S„. 
If  the  series  is  convergent,  its  sum  is  represented  by  S,  which  is 
defined  by  the  relation 

lim.  (.S„)n=x  =  /S. 

If  the  series  is  convergent,  the  series  beginning  with  the 
(w  +  l)th  term  is  a  convergent  series  and  has  a  sum,  repre- 
sented by  Rn,  which  is  called  the  Residue  and  defined   by  the 

relation 

S==S^  +  R,,   or   R,  =  S-S„. 

573.  From  the  above  discussion  and  definitions  it  is  seen  that 
the  expression 

Ui  +  u^  +  Us-l h«„H 

represents  a  definite  number  S  only  when  the  series  is  convergent. 
Since  a  convergent  series  has  a  definite  sum,  it  can  be  used  in 
demonstrations,  but  divergent  and  oscillating  series  can  be  so 
used  only  under  special  restrictions.  Hence,  it  is  important  to 
discover  principles  by  which  the  convergency  of  infinite  series 
may  be  tested. 

ELEMENTARY  PRINCIPLES 

574.  The  ratio  of  the  (n  -\-  l)th  term  of  a  series  to  the  nth  is 
called  its  Ratio  of  Convergency. 

In  a  geometrical  series  the  ratio  of  convergency  is  constant,  but 
in  general  the  ratio  of  convergency  of  a  series  is  variable,  increas- 
ing or  decreasing  with  the  number  of  terms. 

In  1  +-4--H 1 1 1 1-  •••,  the  ratio  of  convergency  is  the  con- 

2     4     8  2"-i     2" 

stant  ^.     In  1  +  3  +  9  +  27  +  — h  3»-i  +  3»  +  •••,  the  ratio  of  convergency  is 
the  constant  3. 

In  1  +  3  +  5  +  7H f-(2n-l)  +  (2n  +  l)H ,   the  ratio  of  conver- 
gency is  the  variable  ratio      ^  "^    ,  which  decreases  as  n  increases. 
2  w  —  1 

575.  The  convergence  or  divergence  of  a  series  may  often  be 
discovered  by  examining  its  ratio  of   convergency,  or  by  com- 


CONVERGENCY  OF  SERIES  499 

paring  the  series  with  an  auxiliary  series  whose  nature  as  con- 
vergent or  divergent  is  already  known. 

576.  Principle  1.  —  The  nature  of  a  series  as  convergent, 
divergent,  or  oscillating,  is  not  changed  by  prefixing  or  removing 
a  finite  number  of  terms. 

For  the  sum  of  a  finite  number  of  terms  is  a  finite  or  fixed  number. 
Hence,  if  the  sum  of  the  series  is  finite,  or  infinite,  or  oscillating,  when  this 
sum  has  been  increased  or  diminished  by  a  fixed  number,  it  will  still  be 
finite,  or  infinite,  or  oscillating,  as  the  case  may  be. 

577.  Principle  2.  —  A  series  of  positive  and  negative  numbers 
is  convergent,  if  the  series  formed  by  making  the  terms  all  positive  or 
all  negative  is  convergent. 

For  the  effect  of  changing  the  signs  of  some  of  the  terms  of  a  convergent 
series  of  positive  numbers  or  of  negative  numbers  is  to  decrease  the  numeri- 
cal value  of  the  sum  of  the  series. 

578.  Principle  3.  —  If  each  term  of  a  given  series,  or  each  term 
after  any  finite  number  of  terms,  is  numerically  less  than  the  corre- 
sponding term  of  an  absolutely  convergent  series,  the  given  series  is 
absolutely  convergent ;  if  each  term  is  numerically  greater  than  the 
corresponding  term  of  a  divergent  series  of  numbers  having  the  same 
sign,  the  given  series  is  divergent. 

Thus,  since  l-}-i  +  J  +  ^+-"  is  an  absolutely  convergent  series,  1  -f- 
i  +  i  +  i+"'  is  a  convergent  series.  But  though  1— ^  +  ^  —  ;!  +  •••  is  a 
convergent  series,  it  does  not  follow  that  l  +  |-f-f  +  f  +  '-'  is  a  conver- 
gent series,  because  the  convergent  series  used  as  a  standard  of  comparison 
is  not  absolutely  convergent. 

Again,  since  l-f^  +  l  +  f-t-^-t-  "-is  a  divergent  series,  1  +  \+{\  +  \) 
+  (i  +  J  +  4  +  i)  +  (i  +  •••  +  A)  +  •••  is  a  divergent  series ;  that  is,  1  +  i 
-f^-|-J  +  ^-|-^  +  |^  +  |  +  >"  is  a  divergent  series. 

579.  Principle  4.  —  If,  for  all  values  of  n  greater  than  a  certain 
finite  value,  the  ratio  of  convergency  of  a  given  series  of  positive 
numbers  is  less  than  the  ratio  of  convergency  of  a  convergent  series 
of  positive  numbers,  the  given  senes  is  convergent;  but  if  the  ratio 
of  convergency  is  greater  than  that  of  a  divergent  series  of  positive 
numbers,  the  given  series  is  divergent. 

Thus,  in    1-1-1  H 1 \ f-  •••  the  ratio  of  convergency  is  a 

variable  having  the  successive  values  1,  i,  ^,  i,  ••••     It  is  observed  that  the 


500  CONVERGENCY  OF  SERIES 

ratio  of  convergency  of  the  given  series  is  smaller,  after  the  third  term,  than 
that  of  the  series  I  +  ^  +  \+  ^  +  •■■,  which  is  known  to  be  convergent. 
Hence,  the  given  series  is  more  rapidly  convergent  than  the  latter. 

The  series  1  +  2  •  2  +  3  •  2"^  +  4  •  2"  +  •••  is  more  rapidly  divergent  than 
the  series  1  +  2  +  2^  +  2^  +  .... 

580.  Principle  5.  —  If  the  ratio  of  the  corresponding  terms  of 
two  series  is  finite  for  all  values  of  n,  and  if  the  terms  of  both  series 
are  jwsitive,  then  both  series  are  convergent  or  both  are  divergent. 

The  above  principle  may  be  established  as  follows : 

Let  Ml,  M2,  Ms,  •••,  M„,  ■••,  and  vi,  Vo,  Vs,  •••,  v„,  •••  be  the  corresponding 
terms  of  two  series  beginning  with  the  first  or  after  any  finite  number  of 
terms.  Let  these  terms  be  such  positive  numbers  that  the  ratio  m„  :  v„  is 
finite  for  all  values  of  n. 

Let  —  =  n,  —  =  r2,  —  =  /"s,  •••,  »m  being  the  greatest  and  r„  the  least  of 

Vi  V2  Vs 

all  the  ratios. 

Then,  Mi  +  u^  +  us  +  •••  =  nui  +  r^vo  +  rzva  +  •••, 

which  is      <  Tmivi  +  v^  +  vs  +  •••),  but  >  Vp  («i  +  172  +  ^3  +  •••)• 

.-.  — lies  between  »•„  and  r„,  and  hence  is  equal  to  a  finite 

Vi  +  V-2  +  VS  +  -  m  PI 

number,  as  C. 

.-.    Ml  +  M2  +  t(3  +   •••  =C{Vi  +  V2  +  V3  +  •••)• 

Hence,  if  the  sum  of  the  terms  of  one  series  is  finite,  the  sum  of  the  terms 
of  the  other  is  finite  ;  and  if  the  sum  of  the  terms  of  one  series  can  become 
infinite,  the  sum  of  the  terms  of  the  other  can  become  infinite. 

Hence,  both  series  are  convergent  or  both  are  divergent. 

TESTS  OF  CONVERGENCY 

581.  First  Test. — An  infinite  geometrical  series  is  absolutely 
convergent  if  its  ratio  is  numerically  less  than  1,  oscillating  if  its 
ratio  IS  —1,  divergent  if  its  ratio  is  1  or  numerically  greater  than  1. 

The  above  test  may  be  established  as  follows : 

By  §  370,  the  sum  of  the  geometrical  series  a,  ar,  ar^,  •••  ar"-^  (1) 

is  5„  =  <1^U!0.  (2) 

1  —  r 

1.    Let  r  be  numerically  less  than  1. 

Then,  since  as  n  increases  indefinitely,  r»  =  0,  (2)  becomes 

\im.Sn=S  =  -^^, 
1  —  r 

a  finite  value,  since  r  is  a  constant  finite  ratio  and  either  positive  or  negative. 

Hence,  in  this  case  the  series  is  absolutely  convergent. 


CONVERGENCY  OF  SERIES  501 

2.  Let  r  =  -  1. 

Then,  (1)  becomes  a,  —  c,  a,  —  a,  •••, 

the  sum  of  the  first  n  terms  of  which  oscillates  between  a  and  0. 
Hence,  in  this  case  the  series  is  oscillating. 

3.  Let  r  be  equal  to  or  numerically  greater  than  1. 

Then,  by  taking  n  great  enough,  the  second  member  of  (2)  may  be  made 
numerically  greater  than  any  assignable  number. 
Hence,  in  this  case  the  series  is  divergent. 

582.  Second  Test. — A  series  is  absolutely  convergent  if,  as  n 
increases  witJiout  limit,  the  limit  of  the  ratio  of  convergence  is  numeri- 
cally less  than  1,  but  divergent  if  the  limit  of  the  ratio  of  convergency 
is  greater  than  1. 

When  the  ratio  of  convergency  approaches  1  as  a  limit,  the  test 
fails  and  other  tests  must  be  applied. 

The  above  test  may  be  established  as  follows : 
Let  lim.'^^^  =  Z. 

1.  When  I  is  numerically  less  than  1. 

Let  r  represent  any  fixed  number  numerically  less  than  1  but  numerically 
greater  than  I.     Also  let  m  be  any  finite  positive  integer. 
Then,  numerical  values  only  being  regarded, 

Mm+l     .  „   Mm+2     .  „    Mm+3  ^  „   Mm+4     .  „ 
Um  Um+x  Um+2  «m+3 

whence     M„+i  <  U^r,  Mm+2  <  "m+ir,  Mm+a  <  Mm+2»',  Mm+4  <  ?<m+3r,  •-, 

or  Um+1  <  UmV,  M„+2  <  UmT^,  U^+S  <  Mm»^,  Mm+4  <  ?<m»^,   •••  ; 

that  is,  on  and  after  a  finite  number  of  terms  each  term  of  the  given  series  is 
numerically  less  than  the  corresponding  term  of  an  infinite  geometrical  series 
whose  ratio  r  is  numerically  less  than  1. 

Hence,  by  the  first  test  and  Prin.  3,  the  series  is  absolutely  convergent. 

2.  When  I  is  numerically  greater  than  1. 
Then,  as  in  1  it  can  be  shown  that 

nm+l>Umr,   «m+2>Mm»^,   «m+3  >  "m?^,   Mm+4  >  Mm^*,    •••  J 

that  is,  on  and  after  a  finite  number  of  terms  each  term  of  the  given  series  is 
numerically  greater  than  the  corresponding  term  of  an  infinite  geometrical 
series  whose  ratio  r  is  numerically  greater  than  L 

Hence,  by  the  first  test  and  Prin.  3,  the  series  is  divergent. 


502  CONVERGENCY  OF  SERIES 

The  failure  of  the  second  test  is  illustrated  as  follows 

In  each  of  the  series  l  +  i  +  i  +  i+ h-H 1 and  1  _  ^  +  ^  _ 

11 

i  +  •••  +  (—  1)"~^-  4-  (—  1)" h  •••,  the  ratio  of  convergency  is  numeri- 

n  n  +  1 

cally  equal  to  — - —    Though  this  ratio  is  numerically  less  than  1  for  finite 

n  +  1 
values  of  n,  it  approaches  1  as  a  limit  as  ji  is  increased  without  limit.  Since 
the  limit  of  the  ratio  of  convergency  is  neither  less  nor  greater  than  1 
numerically,  the  second  test  is  not  applicable.  In  fact  the  first  series  is 
divergent,  as  illustrated  under  Prin.  3,  and  the  second  is  convergent,  by 
the  third  test,  which  follows. 

583.  Third  Test.  —  If  the  terms  of  a  series  are  alternately 
positive  and  negative,  and  each  term  is  numerically  greater  than 
the  following  term,  the  series  is  convergent,  provided  the  limit  of 
the  nth  term  is  zero. 

The  above  test  may  be  established  as  follows : 

Let        U1-U2  +  US-  Ui-\ +  (-  1)»-1m„  + M2»  +  U2n+1 

be  the  series,  and  let       mi  >  «2  >  ms  >  •  ■  •  >  "n  > 0,  so  that  un=0. 
The  sum  of  an  even  number  of  terms  may  be  written 

Ssn  =  («1  -  «2)  +  ("3  -  M4)  +  •••  +  (M2»-1  -  M2n),  (1) 

or  San  =  Ui-  («2  -  «3)  -  (M4  -  Ms) (M2n-2  -  «2n-l)  -  M2».       (2) 

Since  Mi>t*2>«3>  •••,  each  group  in  (1)  and  (2)  is  a  positive  number. 
Therefore,  by  (2),  <S'2„  <  mi  ;  and,  by  (1),  the  sums  *S'2,  St,  Ss,  •-,  Szn 
are  positive  and  increase  with  n. 

Hence,  lim.  (S2n)n^^  lies  between  0  and  ui. 

The  sum  of  an  odd  number  of  terms  can  be  written 
S2n+i  =  S2n  +  W2n+i,  a  positive  number, 

or  S2n+1  =  «1  -  (M2  -  Us)  -  (M4  -  Ms) (M2n  -  M2n+l), 

which  is  less  than  ui. 

Hence,  11m.  (/S^2n+i)n==x>  also  lies  between  0  and  «i. 

Since  the  limit  of  the  nth  term  is  zero, 

lim.  (M2n+l)n=^  =  0 ; 

that  is,  lim.  (52„+i  -  S2n)n^  =  0. 

Hence,  S2n  and  S2n+i  approach  a  common  limit  and  the  series  is  convergent. 

Note.  — If  the  rath  term  approaches  a  constant,  not  zero,  as  a  limit,  say 
lim.  u„  =  a,  Sin+i  and  S2n  approach  limits  that  differ  by  o,  and  the  series  is 
oscillating.  Thus,  |-|  +  |_f +...-|-(_i)«-i^+l  +  ...  jg  an  oscillating  series, 
because  lim.  (*S'2»+i  -  /S2»)„=«,  =  lim.  («2»f  i)»i»  =  1. 


CONVERGENCY  OF  SERIES  503 

584.  Effect  of  grouping  or  rearranging  the  terms  of  a  convergent 
series. 

The  sum  S  of  an  absolutely  convergent  series  is  not  changed  by 
grouping  or  rearranging  its  terms  in  any  manner,  for  by  taking  a 
sufficient  number  of  groups  of  the  derived  series  all  the  terms  of 
Sn  may  be  included,  and  the  sum  of  m  groups  approaches  S  as  S^ 
approaches  S. 

Thus,  §371,  1-^+1  _j  +  ^_^ij  +  ...  =  |j 

also,  §  371,  (1  _  ^)  +  (i  _  ^)  +  (3I5  _  ^^)  +  ...  =  I ; 

also  (1  +  J  _  I)  +  (^^  +  5»^  _  I)  +  (5^,  +  ^^._  ji^)  +  ... 

=  (5_n      /A_1N      /5_1\      .. 

4  V        42     44^      )     2V        22     2*^      } 

§371,  =Kil)-KI)  =  l- 

But  when  a  convergent  series  is  not  absolutely  convergent,  that 
is,  when  its  convergence  depends  partly  upon  the  signs  of  its 
terms,  grouping  or  rearranging  its  terms  may  change  the  sum  of 
the  series  or  even  cause  the  series  to  become  divergent. 

Such  series  are  said  to  be  conditionally  convergent 

Thus,  i_i  +  ^_i  +  ...  +  (_l)«-il4.... 

is  a  conditionally  convergent  series.     Denote  the  sum  by  S. 

Then ,  6^  =  1  -  i  +  i  -  i  +  i  -  i  +  ^  -  J  +  i  -  T^s  +  ^ij  -  ,35  +  . . . ; 
but  (1  +  I  _  i)  +  (1  +  I  _  ^)  +  (J  +  ^  _  I)  +  ... 

=  S-t\S  =  \S.     That  is,  by  this  method  of  grouping  the  sum 

has  been  increased  one  half. 

The  sum  of  this  series  may  be  changed  at  will  by  suitably  grouping  and 
reaiTanging  the  terms. 

By  grouping  similarly  the  terms  of  the  conditionally  convergent  series 
1  —  y/\  +  V|  —  y/\  +  •••,  the  resulting  series  can  be  shown  to  be  divergent. 


504  CONVERGENCY  OF  SERIES 

685.  The  standard  auxiliary  series  for  determining  the  con- 
vergency  of  series  is  the  geometrical  series  l-\-ar-\-  a>•^-|-ar^^ . 

The  next  in  importance  is  the  series  ■::-  +  w\ 1 1-  •••• 

Ip     2^     3*     4^ 

The  series  :i-  +  i^+:^  +  -j-  +  "-  is  absolutely  convergent  when  p 

is  positive  and  greater  than  1 ;  otherwise  it  is  divergent. 

The  proof  is  as  follows: 

1.  Let  p  >  1.     Then,  the  first  term  is  equal  to  1 ;  the  sum  of  the  next  two 

terms  <  — | — ,  or  — ;  of  the  next  four  terms  <  — | i h  — ,  or  —  ; 

2p     2p  2p  4^      4p     4p     4p  4p 

of  the  next  eight  terms  <  —  ;  etc.     Hence,  after  the  first  term,  the  terms  of 

the  series,  grouped  as  shown,  are  less  than  the  corresponding  terms  of  the 
geometrical  series 

2p      iP      8p 

which  is  absolutely  convergent  by  the  first  test. 

Hence,  by  Prin.  3,  the  given  series  is  absolutely  convergent  when  jp  >  1. 

2.  Let  p  —  1.     Then,  the  series  may  be  written 

Since  each  of  the  grouped  terms  is  greater  than  the  corresponding  term  of 
the  divergent  series  1  +  |  -f-  (J  +  ^)  +  (i  +  |  +  |  +  |)  +  ...,  by  Prin.  3  the 
given  series  is  divergent  when  p  =  1. 

3.  Let  p<l. 

It  is  evident  that  the  series  is  more  rapidly  divergent  than  when  p  =  \, 
especially  if  p  is  negative. 

Examples 
1.    Find  the  first  four  terms  and  represent  the  sum  of  the 

series  whose  nth  term  is  (—  1)""^—  •    Test  the  convergency  of  the 
series.  '— 

Solution 

Substituting  1,  2,  3,  4,  ...,  and  m  +  1  successively  for  n  in  (-l)»-i— , 

Ira 


[4         [4'     ""*      "        '  \n4-\      '       '  |ra  +  l 

12     [3     [4  ^       ^      [n      ^  [n±l 


CONVERGENCY  OF  SERIES  505 


Ratio    of  convergency  =  ^^^^  =  ^--. — ^  x 


\n  _i 


M„        I  w  + 1        (- l)''-i       tt  +  1  n  +  l 

which  is  numerically  less  than  1  and  approaches  0  as  a  limit  as  n  increases 
without  limit. 

Hence,  2d  test,  the  series  is  absolutely  convergent. 

2.  Apply  the  third  test  to  the  series  of  example  1. 

3.  Test  the  convergency  of  the  series  of  example  1  by  com- 
parison with  a  geometrical  series. 

Solution 
The  given  series  may  be  written 

i-l  +  J L_  + \ I +  .... 

2     23     2.3.4     2.3.4.5     2.3.4.5.6 
The  infinite  geometrical  series  whose  ratio  is  —  J  may  be  written 

l_l  +  _i 1_  +  --J 1 +  .... 

2      2.2      2.2.2      2.2.2.2      2.2-2.2.2 

Since  each  term  of  the  given  series  after  the  second  is  numerically  less 
than  the  corresponding  term  of  an  infinite  decreasing  geometrical  series,  and 
since,  §  581,  every  infinite  decreasing  geometrical  series  is  absolutely  con- 
vergent, by  Prin.  3,  the  given  series  is  absolutely  convergent. 

4.  Compare  the  series  of  example  1  with  the  auxiliary  series 

14-1  +  1  +  1  +  1+1  +  =.. 
li.'^2^     3^     4^     5"     6^         • 


Solution 
Let  p  =  2.     Then,  the  auxiliary  series  becomes 

1     I     1     I     1     I     1     ,     1     ,     1     I 
1.12.23.34.46.56.6 

The  series  in  Ex.  lis  1-1  +  1-1  +  1-1  +  ... 


1      1.2      12. 3      1.2-3.4      1.2.3.4.5      1.2.3.4.5.6 

each  term  of  which  after  the  third  is  numerically  less  than  the  corresponding 
term  of  the  auxiliary  series. 

Since.  §  585,  the  auxiliary  series  is  absolutely  convergent  when  p  =2,  the 
given  series  is  absolutely  convergent. 


606  CONVERGENCY  OF  SERIES 

5.    Discuss  the  convergency  of  the  series 

Solution 

3x  =  M2  =  (2  . 2  -  l)x2-i ;  5 a;2  =  M3  =  (3  . 2  -  l)x8-i ;  7  x^  =  M4  =  (4  •  2-  l)x*-i. 

Similarly,  «„  =(n  •  2  —  l)a?*-i  =  (2  n  —  l)x»-i,  and  substituting  n  +  1  for 
n,  M„+i  =  [2(n  +  1)  -  ljx(»+i)-i  =  (2  n  +  l)x» 

.    M„+i^    (2n+l)a^  ^271  +  13.^/1  ■       2      \ 
M„       (2n-l)x»-i     2n-l         V       2n-lj' 

•■■  ""■  ["^J-. = «"•  [(' + 2^1)'=]^. = ('  +  ">"^ = "^- 

Hence,  2d  test,  the  series  is  absolutely  convergent  when  x  is  numerically 
less  than  1,  divergent  when  x  is  numerically  greater  than  1,  and  the  test  fails 
when  X  is  numerically  equal  to  1. 

It  is  easy  to  see,  however,  that  the  series  is  divergent  when  x  =  1,  since 
1  +  3  +  5  +  7+ •••is  plainly  divergent.  Finally,  when  x  =  —  1,  the  series 
becomes  1  —  3  +  5  —  7  +  9  —  11  +  •••.  The  sum  of  2  terms  is  —2;  of 
3  terms,  +  3 ;  of  4  terms,  —  4  ;  of  5  terms,  +  5  ;  of  n  terms,  (—  l)"-in. 
Since  by  taking  n  large  enough,  the  sum  of  the  first  n  terms  can  be  made 
numerically  greater  than  any  assignable  number,  the  series  is  divergent. 

Hence,  the  series  is  absolutely  convergent  when  x  is  numerically  less  than 
1,  but  divergent  in  all  other  cases. 


Write  the  first  six  terms  of  the  series  denoted  by 

Find  the  ratio  of  convergency  of 

10.  1+3  +  5  +  7  +  ". .  12.   l_f+5_7  +  .... 

11.  l-^  +  |_J^+....  13.    l_2x  +  3ar^-4a;3+.... 

Discuss  the  convergency  of  the  following  series : 

15.  1+I  +  I  +  I+....  ^  2      2^     2      ^ 

16.  l--i  +  i-|  +  -.  20.    ^^  +  ^_^^  +  ^^_^+.... 

17.  l+f  +  |  +  i+"-.  ^     2  . 2     3  •  2=^     4  .  2^ 

18.  |+f  +  |  +  T'6  +  -.  '  3    "^    3=^    "^    33    "^  " 

1111 
Vl  +  1      V2  +  I      V3  +  1      V4  +  1 


CONVERGENCY  OF  SERIES  607 

23.  l-2ic  +  3a^-4a^  +  5a;*-6ar*4-".. 

24.  l  +  2a;  +  4x'2+6ic3  +  8a;<  +  10ar'+--. 

25.  l  +  ia;  +  |a^+iVa^  +  rr^'  +  T^^  +  -- 

1-2     2-3     3.4^4.5 
29.   a  +  {a  +  d)r  +  {a  +  2 d)!^ -\ \-{a  +  {n~ l)d]r"-^  +  .-. 

586.   Convergency  of  the  binomial  series. 

It  has  been  proved  in  §  452  that  when  w  is  a  positive  integer 
the  series 

a"  +  na"-^a:  +  ^^^^a--V  +  ''^^~^]^^3~^^a»-V+--     (1) 

terminates  and  is  equal  to  (a  +  xy.  It  will  be  shown  in  a  later 
chapter  that  if  (a  4-  xy  can  be  expanded  into  a  series  of  ascend- 
ing powers  of  x  when  n  is  negative  or  fractional,  the  series  is 
infinite  and  has  the  form  of  (1).  It  is  important,  then,  to  inquire 
for  what  values  of  x,  if  any,  this  infinite  series  is  convergent. 
The  (r  +  l)th  and  the  rth  terms,  respectively,  are 

n{n-V) .-  (n  -  r  +  2)(n  -  r  +  1)  ^„-.^ 
1.2...(r-l)r 

^^•^  1.2...(r-l)  ' 

and,  consequently,  the  ratio  of  convergency  of  the  series  is 

/n-r  +  iy^^    /n  +  l_j\x 
\       r       Ja         \    r  Ja 

in  which  n,  x,  and  a  are  constants  and  r  is  a  variable,  expressing 

the  variable  number  of  terms,  1, 2, 3,  •••,00.    Hence,  as  r  increases 

71  4- 1 
without  limit,  approaches  the  limit  zero,  and 

r 

iim.[('i±i-iYl    =(o-i)?=-5. 


a»ri 


508  CONVERGENCY  OF  SERIES 

Kow is  numerically  greater  or  less  than  1,  according  as  x 

is  numerically  greater  or  less  than  a. 

Hence,  by  the  second  test,  the  binomial  series  developed  from 
(a  +  xy  is  absolutely  convergent  ivhen  x  is  numerically  less  than  a, 
and  divergent  when  x  is  numerically  greater  than  a. 

The  second  test  fails  when  x  is  numerically  equal  to  a. 

When  x  =  a,  (1)  may  be  written 

_  1.2  1-2.3  J  ^  ^ 

When  x  =  —  a,  (1)  may  be  written 

a.[l  -  n  +  ^^^  -  "  ("  -  W-'g-  ^^  +  ■■■]■  (3) 

The  nature  of  the  series  in  (2)  and  (3)  depends  upon  the  value 
of  n.  The  full  discussion  of  these  cases  is  beyond  the  scope  of 
this  book. 

In  ChrystaVs  Algebra,  Part  II,  page  131,  it  is  shown  that  (2)  is  absolutely 
convergent  when  n  is  positive,  conditionally  convergent  when  n  lies  between 
0  and  —  1,  oscillating  when  n  =  —  1,  and  divergent  when  n<  —  1 ;  also  that 
(3)  is  convergent  when  n  is  positive,  and  divergent  when  n  is  negative. 

587.   Sum  and  difference  of  two  convergent  series. 

Let  ?7„  =  rtj  +  ^2  +  1*3  +  • '  •  +  Un 

and  Vn=Vi  +  V2  +  Vs-] h  v„ 

be  the  sums  of  the  first  n  terms  of  two  convergent  series  whose 
sums  are  CTand  V,  respectively.     Then,  §§  568,  423,  as  n  =  oo, 

Therefore, 

(ui  +  Vi)  +  (u2  +  V2)  +  (us  +  Vs)-\ 1-  (m„  +  v„)  H 

is  a  convergent  series  whose  sum  is  U+V. 

Hence,  if  the  corresponding  terms  of  ttvo  convergent  series  are 
added,  the  residting  series  is  convergent,  and  its  sum  is  equal  to  the 
sum  of  the  sums  of  the  two  given  series. 

The  same  principle  applies  when  one  convergent  series  is  to  be 
subtracted  from  another,  term  by  term,  for  this  operation  is  equiv- 
alent to  adding  the  convergent  series  formed  by  changing  the 
signs  of  all  the  terms  of  the  subtrahend. 


UNDETERMINED   COEFFICIENTS 


588,  Coefficients  assumed  in  the  demonstration  of  a  principle 
or  the  sohition  of  a  problem,  whose  values,  not  known  at  the 
outset,  are  to  be  determined  by  subsequent  processes,  are  called 
Undetermined  Coefficients. 

To  expand  (x  —  l)(x  +  l)(x  —  2)  without  actual  multiplication, 

put  (x-l){x  +  l)(x-2)  =x«-\- Ax^-\- Bx+  C,  (V) 

and  determine  A,  B,  and  C  by  means  of  the  fact  that  (1)  is  an  identity  and 
must  be  true  for  all  values  of  x. 

By  (1),  when  a:  =  0,  2=0; 

whenx  =  l,  0=\+A  +  B+C; 

when  x  =  -l,  0  =- I  +  A  -  B  +  C  ;  etc. 

By  solving  these  three  conditional  equations,  the  coefficients  A,  B,  and  C, 
undetermined  in  (1),  are  found  to  be  ^  =  —  2,  £  =  —  1,  0=2. 

.:  (x-l)(x  +  l)(«-2)  =x3-2a;2_a;  +  2. 

589.  Let  po  +  Pix  -f  p^x^  +  •  •  •  +  p,.-i«"-'  +  Pn^^  +  ■-,  (1) 
called  a  power  series  in  x,  be  convergent  when  x  =  a,  and  let  it  be 
required  to  find  toward  what  limit  the  sum  S  tends  when  x  =  0. 

The  ratio  of  convergency  of  (1)  when  a;  =  a  is  -^-^  •  a. 

Pn-\ 

Since  (1)  is  convergent  when  x=  a,  the  limit  of  the  ratio  of 
convergency  as  w  =  oo  cannot  exceed  1  numerically,  and  conse- 
quently must  be  numerically  less  than  1  when  x<.a  numerically, 
say  when  x  =  b  (not  zero). 

Hence,  §  582,  p^  +  p^h -{■p^'' +  -  +  p^.^b'^-'  +  p,J)"  +—  (2) 
is  absolutely  convergent.  Therefore,  it  is  possible  to  find  a  fixed 
number,  as  q,  that  is  numerically  greater  than  any  term  of  (2). 

609 


510  UNDETERMINED   COEFFICIENTS 


Take   q>Pib,  q>pj)^,  q>Psb^,  •••,  q>pj>'',  numerically. 
Then,  pi  <  |,  i>2  <  p,  Pa  <  ^3'  ->  P»  <  |.»  numerically. 
Therefore,  by  (1), 
PiX  +p!^  -\-Ps^  +  •••  <  y  +  ^  +  ^  +  •••>  numerically, 


Since  &  is  a  value  of  x  between  a  and  0,  when  a;  =  0  the  series 
in  the  second  member  of  (3)  is  an  infinite  decreasing  geometrical 
series. 

.-.  PiX+p^+p^-\--",  =S—po,  <^-7 ,  or  < -5^, numerically, 

0    b—x  b—x 

which  approaches  zero  as  a  limit  as  x  =  0. 

That  is,  as  x  =  0,  S  —  po  =  0,  or  S  =  Pq.     Hence, 

Principle.  —  If  a  power  series  in  x  is  convergent  for  a  value  of 
x,  not  zero,  it  is  absolutely  convergent  for  any  smaller  value  of  x,  and 
as  x  =  0  the  sum  of  the  series  approaches  the  first  term  as  a  limit. 

590.   JjQt  A  +  Bx  +  Ca?  +  —  =:^  A'  -\-  B'x  +  C'x^  +  ••.,  (1) 

for  every  value  of  x  that  makes  both  series  convergent. 

Then,  for  sxich  values  of  a:,  as  w  =  oo, 
lim.  {A  +  Bx  +  Cx'  +  -)  =  lim.  (A'  +  B'x  +  Ca^+  •..). 

Therefore,  §  589, 
lim.  {A  +  Bx  +  Cx"  +  •..)x=o  =  lim.  (A'  +  B'x  +  C'x"  +  "•)x=0, 
or  A  =  A'.  (2) 

.-.  by  (1),         Bx  +  Cx"  +  -  =  B'x  +  C'x"  +  ... , 
or  x(B+Cx  +  -)=x(B'  +  C'x +'.•).  (3) 

Since  a;  =  0  but  is  not  equal  to  zero,  the  members  of  (3)  may 
be  divided  by  x. 

.-.  B+Cx  +  Dx'-\--  =  B'+Cx-\-D'x'+'-,  (4) 

for  all  values  of  x  that  make  the  given  series  convergent. 

From  (4),  by  the  reasoning  applied  to  (1), 

lim.  (B+Cx  +  Dx'+  •••)x=o  =  lim.  (B'  +  C'x  +  D'x^  +  ...)a;=o ; 
that  is,  B  =  B',  whence  x (C  +  Dx -\-  —)  =  x(C  +  D'x  +  ..•). 

Similarly,  C=C',  D  =  D',  etc.     Hence, 


UNDETERMINED   COEFFICIENTS  511 

Principle  of  Undetermined  Coefficients.  —  If  two  series 
arranged  according  to  the  ascending  powers  of  x  are  equal  for  every 
value  of  X  that  makes  both  series  convergent,  the  coefficients  of  the 
like  powers  of  x  are  equal  each  to  each. 

Since  lim.  {A  -f-  Bx  +  Ca^  +  •••)a;=o  =  ^  also  when  the  series  is 
finite  (§  423),  the  Principle  of  Undetermined  Coefficients  applies 
when  one  or  both  of  the  series  are  finite. 

591.  An  algebraic  expression  is  said  to  be  developed  when  it 
is  transformed  into  a  series  whose  sum,  if  it  is  finite,  or  the  limit 
of  whose  sum,  if  it  is  infinite,  is  equal  to  the  given  expression. 

DEVELOPMENT  OF  FRACTIONS 
Examples 

592.  1.    Develop  the  fraction  - — — — ^« 

^  l  +  x  +  ar^ 

Solution 

The  first  term  of  the  development  is  evidently  1  -r-  1,  or  1 ;  and  since  the 
denominator  is  not  exactly  contained  in  the  numerator,  the  development  is 
an  infinite  series  beginning  with  1  and  proceeding  according  to  ascending 
powers  of  x. 

To  determine  the  coefBcients  of  the  various  powers,  assume 

^  ^^'^    =  \  +  Ax  +  Bx"^  +  Cx3  +  Dx*  +  ...  (1) 

1  +  X  +  x^ 

true  for  all  values  of  x  that  make  the  second  member  a  convergent  series. 
Clearing  of  fractions  and  collecting  terms, 


l+2x=l  +  ^|x  +  -B 
+  1        +^ 


X2+   C 

+  B 

+  A 


x^  +  D 
+  C 
+  B 


X*  +  ... 

+  •••  (2) 

+  - 


+  1 

The  first  member  may  be  regarded  as  the  infinite  series, 

1  +2x  +  0x2  +  0x8  +  ..., 

which  has  a  definite  sum  for  all  values  of  x,  while  the  second  member  is  an 
infinite  series  having  a  definite  sum  for  such  values  of  x  as  make  the  series 
assumed  in  (1)  convergent. 

Therefore,  since  (2)  is  true  for  all  values  of  x  that  make  the  assumed 
series  convergent,  by  the  principle  of  undetermined  coefficients  the  coefficients 
of  like  powers  of  x  in  (2)  may  be  equated. 


512  UNDETERMINED   COEFFICIENTS 

Hence,  A-\-\=2;  .-.  A  =  \; 

JS  +  ^  +  l  =  0;  ..  S  =  -2; 

C+5  +  ^  =  0;  .-.  C=l; 

Z)  +  C  +  B  =  0 ;  .-.  i)  =  1 ;  etc. 

•        1  +  2  «  .  =  ]  +  x  _  2  x2  +  x3  +  X*  -  .... 


\  +  X-\-  X 

The  fraction  may  be  developed  also  by  division. 
2.   Develop  the  fraction  —  ^  +  2x'^ 

Solution 

Q 

Since  the  first  term  of  the  quotient  is  evidently  — ,  or  2  aT", 

surae  2  -  a:  +  2  x"  ^  ^^.^  _|_  j^^.^  +  (7  +  ^a;  +  ^x2  +  — . 

x^  —  2  x^ 

Clearing  of  fractions, 


2-x  +  2x2  =  ^+     B\x+     C 
-2a\    -IB 


X2+       i?|X3+       JFiX*  +  ' 

-2C       -2D 


Equating  the  coefficients  of  like  powers  of  x,  §  590,  and  solving, 
^  =  2;5  =  3;  C=8;Z)=16;^  =  32;  etc. 

.        .^  —  X  +  2  X     _  o  ,v.-2     I     Q  ,>.-l      1     Q     1     1ft«     1     00/>-2     1 


X2  -  2  X3 

Develop  to  five  terms : 

^-     1-0." 

10. 

l-2a;-a^ 

17. 

1     l  +  3a; 

11. 

x  —  o^ 

18. 

1  +a; 

\  +  2x-y? 

5          1 

12. 

\-x 

19. 

'   l-2a; 

6        3 

13. 

1 

20. 

^-    1-x 

\~X  —  '3? 

7         1 

14. 

2  +  a;-2a^ 
l-a;  +  2a:2 

21. 

1  —  ace 

Q     2  +  3ar2 
'    1-2.'b2 

15. 

a^  +  a^ 
l-2a;  +  a^ 

22. 

4rc-3a^ 
■     l  +  2a; 

16. 

l-2a; 

^- + -i?  -ir  x^ 

23. 

2-5a; 

2x-Q^ 

l  +  a;  +  a^ 

a;  +  cc^  +  ic* 

1 

a  — ic 

1 

an- a? 

a 

1-a; 

a 

h  —  X 

c 

CIX 


UNDETERMINED   COEFFICIENTS  513 

DEVELOPMENT  OF  SURDS 
Examples 


693.  1.  Develop  the  expression  Va  -|-  a;  by  the  use  of  unde- 
termined coefficients. 

Solution 

Assume  y/a  +  x  =  A  +  Bx  +  Cx"^  +  Dx?  +  Ex^  +  •■•. 

Squaring,  a-\-x-  A^  +  2  ABx  +  B^\x^  +  2  AD\x^  +  C'^lx*  +  •... 

+  2AC\      -\-2Bc\->r2AE\ 
+  2Bd\ 
Equating  the  coefficients  of  like  powers  of  x,  §  590, 

A'^  =  a;   .-.  J  =  Va. 

2AB=\;    .:  B  =  ^. 
2a 

B^  +  2AC  =  0;    .'.   C=  -— . 
8a2 

2AD  +  2BC  =  0;    .:  D  =  -^. 
16  a8 

C^  +  2AE  +  2BD  =  0;    .:  E=- -A^. 

128  a* 

.*.  va  +  X  =  v«    1  -\ h  • —  ). 

V        2a     Ba-'^      16a3     128  a*         / 

The  given  surd  may  also  be  developed  by  the  extraction  of  the  root  indi- 
cated or  by  the  use  of  the  binomial  formula.  But  whatever  the  method  of 
development,  the  series  obtained  is  equal  to  the  surd  only  for  such  values  of  z 
as  make  the  series  convergent. 

Develop  to  five  terms  by  undetermined  coefficients : 


2.    VI  —  X.  8.    VI  + 


X. 


3.  VI  +  2 «.  9.  vr+^T^. 

4.  Vl  +  4  aj.  10.  (1  +  4  a;  +  6  ar^  4- 4  a^  +  a?*)i 

5.  V4-}-a;.  11.  {l-3x-\-3x^-a^i 

6.  Vo^^.  12.  (l+x)i 


7.    ^/a^-x^.  13.    Vl  +  2x  +  3x^  +  i3^  +  .'.. 

ADV.  ALG.  —  33 


514  UNDETERMINED   COEFFICIENTS 

PARTIAL  FRACTIONS 

594.  To  resolve  a  fraction  into  partial  fractions  is  to  separate  it 
into  simpler  fractions  whose  sum  is  equal  to  the  given  fraction. 

It  is  necessary  to  consider  only  proper  fractions.  For  if  the 
degree  of  the  numerator  is  not  lower  than  that  of  the  denomi- 
nator, the  fraction  may  be  reduced  by  division  to  the  sum  of  an 
integer  and  a  proper  fraction. 

If  the  denominator  of  a  fraction  is  composite,  the  fraction  may 
be  resolved  into  an  indefinite  number  of  sets  of  partial  fractions. 
But  the  number  of  sets  is  limited  to  one  by  requiring  that  each 
partial  fraction  shall  be  a  proper  fraction  incapable  of  resolution 
into  simpler  fractions. 

1.    To  select  the  denominators  of  the  partial  fractions. 

Since  the  resolution  of  a  fraction  into  partial  fractions  is  the 
converse  of  the  process  of  uniting  fractions  with  different  denomi- 
nators into  a  single  fraction  having  their  lowest  common  denomi- 
nator, it  is  evident  that  the  denominators  of  the  partial  fractions 
must  comprise  all  the  various  prime  factors  of  the  given 
denominator. 

When  the  prime  factors  of  the  given  denominator  are  all 
different,  it  is  sufficient  to  assume  each  of  them  for  the  denomi- 
nator of  a  partial  fraction.  But  when  any  prime  factor  is 
repeated  in  the  given  denominator,  say  m  times,  there  must  be 
a  partial  fraction  having  the  wth  power  of  this  factor  for  a 
denominator,  for  otherwise  the  L.  C.  M.  of  the  denominators 
would  not  contain  the  mth  power  of  the  factor.  There  may  or 
may  not  be  denominators  that  are  lower  powers  of  the  factor 
than  the  mth,  but  provision  should  be  made  for  such  denomi- 
nators. If  they  do  not  occur  among  the  partial  fractions,  this 
will  be  indicated  by  numerators  that  reduce  to  zero. 

Two  illustrations  of  the  selection  of  denominators  will  be  given. 

Thus  5x^--Sx-2^        _  ^  ^  .  ._^2^  .  _  .^        (1) 

and    ^-26^0^+1^--    ^i_   +-J^    ^      ^«      +_Z^_-.      (2) 


UNDETERMINED   COEFFICIENTS  515 

In  (1)  no  numerator  can  be  equal  to  zero  because  each  denomi- 
nator is  needed  to  produce  the  L.  C.  D.,  (a^  +  1)  (a;  +  3)  {x  +  4). 
But  in  (2)  it  is  possible  that  ^5  or  N^  may  be  equal  to  zero,  since 
the  L.  C.  D.  would  not  be  changed  by  omitting  the  corresponding 
fractions. 

2.  To  assume  expressions  for  the  numerators  of  the  partial 
fractions. 

It  has  been  agreed  that  each  numerator  shall  be  of  lower  degree 
than  the  corresponding  denominator,  and  that  each  partial  frac- 
tion shall  be  of  such  a  form  that  it  cannot  be  resolved  into 
simpler  fractions. 

Therefore,  in  (1)  the  numerators  -^2  and  ^3  must  be  numerals, 
and  the  numerator  N^  may  be  either  a  numeral  or  an  expression  of 
the  first  degree  in  x.  Hence,  the  numerators  of  the  partial  frac- 
tions have  the  form 

5cB2_3a.--24        _Ax  +  B  ^      C      .      D  ,^. 


(a^  +  1)  (a;  +  3)  (a;  4- 4)       x'  +  l       x  +  3      x  +  4. 

T     r    .            5ar'-3a;-24             «-2,       3  4 

In  fact,   — — =  — \- 


(ar^ -I- 1)  (x  4- 3)  (a;  +  4)      a.-^  +  1   '  a- +  3      a; -f  4 

In  (2)  the  numerators  N^  and  iV^  corresponding  to  powers  of 
{x  -f  3),  the  prime  factor  in  the  denominator,  must  be  of  lower 
degree  than  this  prime  factor,  for  otherwise  either  numerator 
could  be  separated  into  two  parts,  one  of  them  a  multiple  of 
{x  4-  3),  which  would  allow  the  fraction  to  be  further  decomposed. 

Hence,  the  numerators  of  the  partial  fractions  have  the  form 

26a;4-18       _Ax  +  B  ,      C      .        I)        ,       E  ... 

"■"  „    1     o'/^    ,     o\2    '     /„    1     o\s'  V    / 


(a^  +  l){x  +  Sy       x^-fl       a; -f  3      (.r  +  3)-      (a! -4- 3)3' 

in  which,  as  special  cases,  either  A  or  B,  but  not  both,  may  be  equal 
to  zero,  and  either  C  or  D  or  both,  but  not  E,  may  be  equal  to  zero. 

In  fact  26a^  +  18       ^      1 1__ 6_, 

'    (x' -\- 1)  (x  +  Sy     af  +  l     (x  +  3)-     (x  +  sy 

Rule.  —  Take  each  prime  factor  of  the  given  fraction  for  a 
denominator  of  a  paTiial  fraction,  and  ivhen  any  j^i'^'^ne  factor 
occurs  m  times  in  the  given  denominator,  use  also  its  second,  third, 
fourth,  •••,  mth  powers  as  denominators. 

Assmne  for  each  numerator  an  expression   tvith   undetermined 


516  UNDETERMINED   COEFFICIENTS 

coefficients  of  a  degree  one  lower  than  the  prime  factw  of  the  cor- 
responding denominator. 

Suppose  that  the  denominator  of  the  given  fraction  is  of  the 
nth  degree  in  x.  Then,  by  the  above  rule,  there  will  be  n  unde- 
termined coefficients ;  and  if  the  assumed  identity  is  cleared  of 
fractions  and  the  second  member  is  then  arranged  according  to 
powers  of  x,  the  second  member  will  be  of  the  (ii  —  l)th  degree 
and  have  n  terms. 

Hence,  by  §  590,  the  two  equal  series  obtained  by  clearing  of 
fractions  will  furnish  n  conditional  equations  from  which  the  n 
undetermined  coefficients  may  be  found. 

Examples 

3 
1.    Resolve into  its  partial  fractions. 

Solution 

3  An 

Assume  that  =  — — 1 ^^—  is  an  identity. 

l-5x  +  6a;2      l_3x      l-2x 

Clearing  of  fractions,  3  =  yi  —  2^x+B  —  3  Bx. 

That  is,  Z  +  a  X  =  A  +  B  -  \2  A  ^  '^  B)  X. 

.-.  §  690,  ^  +  -B  =  3  and  2  ^  +  3  J9  =  0. 

Solving,  ^  =  9  and  JS  =  -  6. 

3  9  6 


Hence, 


l-5a;  +  6x-''     l-3a;     l-2x 


2.    Resolve ; -— -  into  its  partial  fractions. 

1  — 4a;  +  4a^  ^ 

Solution 

Assume  that  :; — ^  ~  ^  ^^     ,  =  —^ —  + is  an  identity. 

\-^x  +  4:X^     \-2x      (l-2a;)2  ^ 

Reducing,  b  —  Qx  =  A  +  B—2  Ax. 

.:  §  690,  ^  +  B  =  5and  -  2  A  =  -  6. 

Solving,  ^  =  3  and  B  =  2. 

Hence,  5-6x     ^_J__^         2 


1  -  4  X  +  4  a;2      l  -  2  a;      (1  -  2  a;)2 


UNDETERMINED   COEFFICIENTS  517 

7 11 7;  -I-  7  a^ 

S.    Resolve '—^ into  its  partial  fractions. 

(1  -  x)\l  +X  +  x") 

Solution 
It  is  evident  that  the  fractions  corresponding  to  the  factor  (1  —  x)*  will  be 

^  ^  and         ^ 


1-x   (i-xy-2  (i-x)^ 

Since  the  factor  (1  +  x  +  x^)  is  quadratic  and  has  no  rational  simple 
factor,  the  numerator  corresponding  to  the  denominator  1  +  x  +  x"'^  may 
have  two  terms  ;  therefore,  assume  that 

7-llx  +  7x'2       _     A  B  C  D-irEx         .j. 


(1  -  X)S(1  +  x  +  X2)         1-x        (1  -  X)2        (1  -  X)3        1+  X  +  X^ 

is  an  identity. 

Then,  7-11x4-7x2  =  ^(1  -  x)2(l  +  x  +  x2)  +  ^(1  -x)(l  +  x  +  x2) 
+  C(H-x  +  x2)  +  (Z)+^x)(l-x)8  (2) 

is  an  identity ;  that  is,  is  true  for  all  values  of  x. 

Since  there  are  five  coefficients,  A,  B,  C,  D,  and  E,  to  be  determined, 
and  since  (2)  is  true  for  all  values  of  x,  by  giving  x  in  succession  each  of 
five  different  values,  five  independent  equations  involving  the  undetermined 
coefficients  may  be  formed,  and  from  these  equations  the  coefficients  may 
be  determined. 

Let  X  =  1  ;  then,  (1  —  x),  (1  —  x)2,  and  (1  —  x)^  reduce  to  0,  and  the 
identity  (2)  becomes 

3  =  3C;  .-.  0=1.  (3) 

Let  X  =  0  ;  then,  7  =  A  +  B+C+B, 

or,  since  C  =  1,  A  +  B-j-  Dr=6.  (4) 

Let  X  =  -  1  ;  then,  2b  =  4A  +  2B+\  +  SD-8E, 
or,  dividing  by  2,  2A  +  B  +  4D-4E-12.  (5) 

Let  X  =  2  ;  then,  13  =  7A-7B  +  7-D-2E, 

or  1A-1B-D-2E  =  Q.  (6) 

Let  X  =  -  2  ;  then,  57  =  27  ^  +  0  S  +  3  +  27  D  -  54  j&, 
or,  dividing  by  0,  3A  +  B-\-'iD-QE  =  %.  (7) 

Solving  the  equations.  (4),  (5),  (6),  and  (7),  we  have,  together  with  (3), 
^  =  2,  5  =  0,   C=l.  Z)  =  4,  E-2. 

Hence,         7  -  11  x  +  7  x^      ^  _2_  ^        l       ^    4  +  2x   . 


(1  -  x)8(l  +  X  +  x2)      1-X      (1  -  x)8      1  +  X  +  xa 


518.  UNDETERMINED   COEFFICIENTS 

X  -\~  2  x^ 

4.  Resolve  -— l- —  into  its  partial  fractions. 

1  —  .^■^ 

Suggestion.  —  Assume  ?-±l^  =  _^  +    B  +  Cx  ^ 
1  -  X^        1  -X      1  +  x  +  x^ 

5.  Resolve  — ,  ^  ~     into  its  partial  fractions. 

ar  —  1 

Suggestion.  —  When  the  numerator  is  not  of  lower  degree  than  the  de- 
nominator, the  numerator  should  be  divided  by  the  denominator  until  the 
remainder  is  of  lower  degree  than  the  denominator.  The  fractional  part  of 
the  resulting  mixed  expression  may  then  be  resolved  into  partial  fractions, 
and  these  may  be  annexed  to  the  integral  part. 

Resolve  each  of  the  following  into  its  partial  fractions : 

—  6x 


2x 

S-Gx  +  x" 

3-\-Ax 

l_|-8a;  +  16iK2 

3a; 

l  +  ar^ 

l-x-ex" 

x  —  x^ 

l-2a!  +  2a^ 

12. 


13. 


8.       "        •  14. 


15. 


10.    -± •^•"^^"^  ,  .  16. 


(x  -  5)3 

x'-S 
x^-l 

2a^  +  9x+ll 

a^  +  4a;-f4 

2  -  6  .T  -f  6  a^ 
l-6a;  +  llar'-6a;3' 

49 


11     3a;-2  ^^     1  +  2a;  4- 3a^  +  ga:^ 


(l-x)(l-2xy  (2_3a;)2(3-a;) 

3a;-2 
(a; -3)=^ 

REVERSION  OF  SERIES 

595.  To  revert  a  convergent  series  of  ascending  powers  of  x  is 
to  express  the  value  of  x  by  means  of  a  series  of  ascending  powers 
of  the  sum  of  the  given  series. 

Let  it  be  required  to  revert  the  series 

y  =  ax  -{- ba^  -{-  cx^  +  dx*  +  •  •  •,  (1) 

in  which  x  has' any  value  that  will  render  the  series  convergent. 
Assume  x  =  Ay  +  By--{-Cf  +  Dy*  -{ .  (2) 


UNDE  TERM  I  NED   COEFFICIEN  TS 


519 


Substituting   this  value  T)f  x  in  (1),  and  dropping  all  terms 
involving  a  higher  power  of  y  than  the  fourth, 


y  =  aAy  +  aB  I  y^  +  aC 

+  bA'\     +2  bAB 
+  cA' 


Since  (1)  is  an  identity,  §  590, 
aA  =  1 ;  .-.  A  = 

aB  +  bA'  =  0;  .:  B  = 

aC+2bAB  +  cA^  =  0;        .:  (7  = 


f  +  aD 
+  bB^ 
+  2bAC 
-\-3cA'B 
+  dA* 


y*  + 


bA^^      b 
a  a? 

-  2  bAB  -  cA^ 
a 


2b'-ac 

«5 


aD+bB^  +  2  bAC  +  3  cA'^B  +  dA*  =  0; 

T)  __  _  g^d  —  5  abc  -f  5  b^ 
~  a' 

Hence, 


5  abc  +  5  6^  4  , 

— r-^ y  + 


1         6    9  ,  2  &^  —  ac  .J      a'd 

«  =  -2/  -  ^r  H ^ — y 


Examples 

1.    Revert  the  series  y  =  x 1 1-  .-. 

^  2      3      4 

Solution.  —  Substituting  in  (8)  1  for  a,  —  ^  for  fe,  |  for  c, 

values  of  the  undetermined  coefficients  in  (2)  are  found  ; 

whence,  x  =  y  +  ^y'^  +  ly^  +  ^y*  -] . 

Revert  the  series  in  the  following  equations : 

2.  y  =  x -i- x^ -\- x^  +  X* -{- ■••. 

3.  y  =  x  —  3x^  +  5x^  —  7x*+  •■: 

4.  y  =  x  +  2x^+3a^  +  Ax*-] . 

5.  y  =  2a;4•3a^4-4a^  +  oa^^ . 

/y2  /yi3  n/A 

6.  y  =  x+  ^-}.5L+.^+.... 
^  2624 

7.  y  =  X  —  3  .T^  +  5  a^. 

8.  y  =  x-2a^  +  2ix^-2x*  +  "'. 


(3) 


^  for  d,  the 


620  BINOMIAL    THEOREM 

9.   Find  the  approximate  value  of  x  to  four  terms  in  the  series 

2~2     12     30     56         ■ 

Solution 

Reverting,         x  =  2(i)  -  K^)^  -  Mh)'  -  i¥V(i)*  -  - 

=  1  ~  i  —  ^  ~"  j-hs  —  ••• 
=  .8189+. 

Find  the  approximate  value  of  x  in  the  following : 

10.  J  =  a;  +  -  +  — +  —  +•••. 
2  6      24     60 

11.  l  =  .-^V^-2^+.... 
5  3       10        7 


BINOMIAL  THEOREM  -  FRACTIONAL  AND 
NEGATIVE  EXPONENTS 

596.    It  has  been  shown  (§  453,  formula  IV)  that  when  n  is 
a  positive  integer, 

(l  +  x)"  =  l+nx.  +  !i(^a-  +  ^(!L^il^^  (1) 

It  is  yet  to  be  proved  that  this  formula  is  true  when  n  is  a 
positive  fraction,  a  negative  integer,  or  a  negative  fraction. 

I.    When  n  is  a  positive  fraction. 

Let  n  =  -,  in  which  jp  and  q  are  positive  integers. 


Then,  §  246,  (1  +  xy  =  ^(1  +  a;)^ 


§452,  =Vl+F«+-.  (2) 

Assume,  §  593,  -\/\+px+  •••  =  A  + Bx +  Gx' -{- ."  (3) 

where  x  may  have  any  value  that  will  make  both  series  convergent. 
Raising  both  members  to  the  gth  power, 

1  +px  +  ...  =\_A  +  iBx-{-Cx^  +  •••)]' 

=  A^->rqA^-'^{Bx-\-Ca?+  •••)  +  •••. 


BINOMIAL    THEOREM  521 

Equating  the  coefficients  of  like  powers  of  x  in  two  terms, 

1  =  J."  and  p  =  qA^-^B) 

P 
whence,  A  =  l,  and  B  —  — 

Substituting  these  values  in  (2)  and  (3), 

^  q 

That  is,  the  formula  is  true  for  two  terras,  when  n  is  a  positive 
fraction. 

II.    Wlien  n  is  negative,  and  either  integral  or  fractional. 

1 


(1 4-  xy 
1 


§  244,  (1  +  x)-'^ 

§  452  and  Case  I, 

1  +  nx  +  ••• 

Dividing  the  numerator  by  the  denominator, 

(1  +  x)-"  —  l  —  nx-] . 

That  is,  the  formula  is  true  for  two  terms  when  n  is  negative 
and  either  integral  or  fractional. 

Therefore,  the  formula  is  true  for  tivo  terras,  whether  the 
exponent  is  positive  or  negative,  integral  or  fractional,  and  the 
coefficient  of  the  second  term  is  n. 

597.   To  find  the  coefficients  of  terms  after  the  second,  assume 

(1  +  xy  =  l  +  nx  +  Ax"  +  Ba^+Cx*+—.  (1) 

Since  only  two  terms  of  the  expansion  of  the  first  member  are 
known,  the  coefficients  A,  B,  etc.,  cannot  be  determined  from  (1) 
in  its  present  form. 

To  find  the  value  of  the  undetermined  coefficients  A,  B,  etc.,  we 
involve  them  in  an  identical  equation,  so  that  the  coefficients  of 
like  powers  of  the  sarae  variable  raay  be  equated  (§  590). 

Since  in  (1)  x  represents  any  number,  put  l-\-{x  +  z)  for  1  +  x. 


622  BINOMIAL    THEOREM 

Then,      (1 -f-aj  +  z)»  =  [l  +  (x  +  z)]" 
by  (1),  =l  +  n{x-\-z)  +  A{x  +  zy  +  B{x  +  zf-\-'" 

=  l+nx  +  Ax^  +  B2(?-\ 

+  {:,i  +  2Ax  +  ^Bx^ +-■)%  +  •■:        (2) 
Since  in  (1)  x  represents  any  number,  and   since   1  +{x-\-z) 

=  {l  +  x)  +  z^{l+x)(l+-^,   put   1  +  :^^  for  1+x. 
Then, 

(ij^x  +  zy  =  {\+xy(\+^y 

by(i),  ■  =a-i-.r(i+.^^+^^.^^^3+-) 

-=  (1 4-  xy  +  ?i  (1  +  xy-h 

+  ^  (1  +  xy-h""  +  B{1+  xy-h^  +  . . ..  (3) 

Therefore,  from  (8)  and  (2),  Ax.  1, 
(1  +  xy  +  n  (1  +  xy-h  +  ^  (1  +  a;)"- V  ^B{l-\-  a;)""  V  +  • .  • 

=  1+  na;  +  Aa?  +  Bx^  -\ \- {n  +  2  Ax  +  ^  Bx^ -\ )z^ 

when  both  members  are  convergent. 

Equating  the  coefficients  of  z, 

§  590,  ?i(l  +  xy-^  =  n  +  2 ^a;  +  3  Ba^  +  — . 

Multiplying  each  member  by  1  +  a;, 

w(l  +  a;)»  =  7t  +  (2^4-w)a;+(3^  +  2^)a:2_,_...^ 

(1)  X  n,  n(l-\-xy  —  n-\-'in?x-\-nAii?-\ . 

Equating  coefficients,       2  A  +  n  =  n^, 
and  3B  +  2.4  =  w^; 

whence,  A  =  ^   ^  ~^  K 

and  .(n-l)(.-2X 

1.2.3 


BINOMIAL    THEOREM  523 

In  like  manner  any  number  of  successive  coefficients  may  be 
found.     Substituting  these  values  in  (1), 

{l  +  xy  =  l  +  nx  +  '^^f^^  +  ^^"  -  ^l%-  ^>a^+  •>•.     (4) 

The  expansion  of  (1  +  a;)"  is  not  the  expansion  for  the  most 
general  form  of  a  binomial,  since  the  first  term  is  1 ;  but  putting 

X 


for  X 
a 


or 


V       ay  a         1-2     a'  1-2.3        a^         '^^ 

\    a    j       a\  1-2 


Multiplying  by  a", 

(a  +  a:)" = a"  +  ««"  'a;  +  ^^^^^  ~/^  a"" V  +  •  •  • .  (6) 

J.    *  ^ 

The  binomial  formula  has  thus  been  proved  true  when  n  is  any 
positive  integer,  any  positive  fraction,  any  negative  integer,  or 
any  negative  fraction,  that  is,  when  n  is  any  commensurable  expo- 
nent, provided  the  second  member  of  the  formula  is  convergent 
when  it  is  an  infinite  series. 

598.  It  has  been  shown  (§  457)  that  the  series  developed  from 
(a+xy  is  infinite  for  fractional  or  negative  values  of  n,  and  (§  586) 
convergent  when  x  is  numerically  less  than  a.  Hence,  in  this 
case  the  true  value  of  (a  +  »)"  may  be  found  to  any  required 
degree  of  accuracy.  The  series  is  divergent  when  x  is  numerically 
greater  than  a,  but  in  this  case  the  true  value  of  (a  +  a;)"  may  be 
found  to  any  reqiiired  degree  of  accuracy  by  expanding  (a;  +  a)", 
for  the  latter  expansion  then  gives  a  convergent  series. 

Thus,  VTOI  is  not  found  from  (1  +  100)i  =  1  +  .50  -  2500  +  •••,  but  from 
(100  +  \)^  =  \{)  -\-  ^  -  ^^  +  ...,  which  approaches  VIoT  as  a  limit. 

When  x  =  —  a,  (a  +  a;)»=0"  =  0 ;  when  x  =  a,  (a  +  a-)"  =  (2  a)", 
the  value  of  which  may  be  found  by  separating  2  a  into  a  binomial 
whose  first  term  is  numerically  greater  than  the  second  and  ex- 
panding by  the  binomial  formula.    Thus,  (5  +  5)^  =  10^  =  (9  + 1)^. 

Exercises  for  practice  will  be  found  on  page  414. 


EXPONENTIAL  AND  LOGARITHMIC  SERIES 


599.  In  a  previous  chapter  the  student  learned  the  use  and 
advantage  of  common  logarithms.  But  the  principles  upon  which 
the  computation  of  the  tables  was  based  had  not  then  been 
established.  Now,  however,  by  the  application  of  the  principles 
of  convergence  of  series,  he  will  be  able,  by  means  of  two  infinite 
series,  called  the  exponential  and  logarithmic  series,  respectively, 
to  derive  a  formula  for  computing  logarithms  in  any  system. 

600.  The  exponential  series. 

The  exponential  series  is  the  development  in  ascending  powers 
of  X  of  the  xth.  power  of  a  certain  constant  base.  The  series  is 
derived  as  follows: 

By  the  binomial  formula,  if  nx  is  commensurable  and  n  is 
numerically  greater  than  1, 

f^      VV"  _  ^    I  *^^  _i_  nx{nx  —  1)      nx{nx  —  1)  {nx  —  2)  .^. 


(2) 


'\2  n^\^ 

When  x  =  \,  (1)  becomes 

(l  I  ^Y  =  l  I  1  I  H^-1)  I  ri{n-l){n-2) 
\        n)  ^     ^      n^^    /^  w3[3  ^ 

By  (2)  and  (1),  since    (l  +  ^X  '  =  (l  +  ^^> 

\  I  1  I  K^-1)  I  n(n-l)(n-2)  T 

_  n'\2  «3|3  ^     J 

_  w    ,       ,  nx(nx  —  1)      nxjnx  —  l){nx  —  2)  ,o\ 

~  i?J2  n3|3  ^^ 

In  (3)  it  is  permissible  to  let  x  have  any  finite  value  while  n 
increases  numerically  without  limit.     For  whatever  the  value  of 

624 


I 


EXPONENTIAL  AND  LOGARITHMIC  SERIES      525 

X,  the  law  of  variation  of  n  as  ?i  =  ±  oo  may  be  so  taken  that  nx 
is  always  commensurable.     Accordingly,  let  n  =  ±  oo . 


Then,  in  (3),  lini. 
lim. 
lim. 


i(n  -  1)' 


,  or  lim.  ( 1  —     )  =  1 ; 


')>(n-l)(n-2)' 


,  or  lim.  (  1  — 


3« 


=  1; 


)ix(nx  —  1) 


,  or  lim.  [x^ |  =  a^ ; 


and  so  on.     Hence,  for  all  finite  values  of  x,  (3)  becomes 


usually  written 


\2  '13 
^^  =  ^  +  ^^1  +  1  + 


(4) 


(5) 


y 


In  (5)  the  base  e  is  a  constant,  1  +  1+  —  +  —  H ,  whose  value, 

I 

as  shown  later,  is  2.7182818  approximately ;  and  the  exponent  x  is 
a  variable,  and  so  may  have  any  finite  value. 

Since  in  e''  the  variable  is  an  exponent,  e""  is  called  the 
exponential  function  of  x,  and  the  series  developed  from  e""  is  called 
the  exponential  series. 

601.  To  derive  a  formula  applicable  to  any  positive  constant 
base  a,  let  log^  a  =  k. 

Then,  a  =  e*, 

and  a"  =  e'"  =  e('°Ke«)''. 

...  by(5),      a^  =  l  +  (log.a)a.  +  (l5S^V^^^^S^V'-..   (6) 

for  all  finite  values  of  x.     This  is  the  Exponential  Formula,  or  the 
exponential  series  when  the  exponent  of  e  is  (log^  a)x. 

602.  In  the  exponential  series  (5)  the  ratio  of  convergency  is 

Iv*.  I  1  Jb  tie  JO 

?<,.       \r      !?•  —  1      r 
in  which  the  exponent  x  is  fiiiite. 

Therefore,  as  r  increases  without  limit,  the  ratio,  of  convergency 
approaches  zero  as  a  limit.  Hence,  §  582,  the  exponential  series 
is  absolutely  converger^  for  all  finite  values  of  the  variable  involved. 


526        EXPONENTIAL  AND  LOGARITHMIC   SERIES 


603.    The  constant  e  is  used  as  the  base  of  a  system  of  loga- 
rithms called  the  Napierian  System*  or  the  natural  system.     Its 

approximate  value  is  computed  from  the  series  l  +  l  +  i-  +  l-| 

as  follows : 


Adding, 


l-fl 

l-^[3  =(1-12)  -3  = 

1h-^  =(1-|3)  -^4  = 

1^[5  =(1^[4)  -^5  = 

1^16  =(1^[5)  ^6  = 

l-^[7  =(1-^(6)  ^7  = 

lH-|8  =(1^|7)  -^8  = 

1-|20  =  (1^|9)  -10  = 
1 -[11  =  (1-110)- 11  = 


=  2.000000000 
=  .500000000 
.166666667 
.041666667 
.008333333 
.001388889 
.000198413 
.000024802 
.000002756 
.000000276 
.000000025 


e  =2.7182818      to  7  places. 


Examples 
From  the  exponential  series  deduce  the  following ; 

1.  i(e^-e-=)  =  x +-  +  -  +  -  +  .... 
^\  J  1 3     [5     [7 

2.  i(e  +  e-i)  =  l  +  l-|.l  +  l+.... 
2^   ^      ^  [2     [4     [6 

3.  ie  =  l+l  +  ^+i  +  .... 
'  |3     [5     [7 

4.  1=2      4      6      8      _ 

e      L§     L?     IJ     [9 

5.  i(e<'4-e-'^)=l--  +  --^V— . 

^^  ^  |2     [4     |6 


*  Logarithms  were  invented  by  Baron  Napier  of  Scotland  in  1614. 


EXPONENTIAL   AND  LOGARITHMIC  SERIES       527 

604.    The  logarithmic  series. 

The  logarithmic  series  is  the  expansion  of  log^  (1  +  a;)  in  ascend- 
ing powers  of  X.     It  is  derived  as  follows : 

By  the  exponential  formula,  when  1  +  a?  is  the  base  and  y  the 
exponent, 

(l4-a.y  =  l  +  [log.(l+a.-)].v  +  ^^°^'^^^-^^^^''^V»-.       (1) 

By  the  binomial  theorem, 

(l  +  ^y  =  l+y.  +  yJl^^  +  H=^!lii^^^+....   (2) 
Equating  the  second  members  of  (1)  and  (2), 

1 + [log.  (1 + .)], + p°g-(^+-)]y + ['°«-(i^+-)]y + ... 

l  +  y.  +  y^^  +  y'^-'^^^-^^^+....  (3) 

Equation  (2)  is  true  when  x  is  numerically  less  than  1  (§  586), 
and  y  is  finite  and  commensurable  (§  597).  Equation  (1)  also  is 
true  under  these  conditions,  for  the  exponent  y  is  finite,  and  when 
X  is  numerically  less  than  1,  the  base  1  +  x  is  positive. 

Therefore,  since  the  two  series  have  the  same  sum  (1  +  a;)*  and 
both  are  absolutely  convergent  for  all  finite  commensurable  values 
of  y,  X  being  numerically  less  than  1,  equation  (3)  is  an  identical 
equation.  Accordingly,  equating  the  coefficients  of  y  in  the  two 
series,  §  590, 

loc  (1  I  x)-x  \-^x'-\  (-^)(-^):^\  (-l)(-2)(-3)   , 
log,(l+a)-x  +  ^    2^+     1.2.3        ^       1.2.3.4       "^  ^     ' 

Simplifying  the  second  member, 

log,(l  +  a.)=a;-|  +  |-|  +  -.  .  (4) 

This  series  is  called  the  Logarithmic  Series.  It  is  absolutely 
convergent  when  x  is  numerically  less  than  1  (§  582),  and  con- 
ditionally convergent  when  a;  =  1  (§§  583,  584), 


628        EXPONENTIAL  AND   LOGARITHMIC  SERIES 

605.    To  compute  a  table  of  natural  logarithms. 

Since  the  logarithmic  series 

log,(l4-a.)=a;-|  +  |-|V...  (1) 

is  not  convergent  for  values  of  x  greater  than  1,  it  cannot  be  used 
vo  find  the  natural  logarithm  of  any  number  greater  than  2. 

Hence,  it  is  necessary  to  modify  (1)  to  make  it  available  for 
computing  the  natural  logarithm  of  any  positive  number,  however 
great ;  and  for  ease  in  computation  it  is  desirable  that  the  series 
obtained  be  rapidly  convergent. 

Substituting  —  x  for  x  in  (1), 

log.(l-a-)=-^-|-|-|*-....  (2) 

Subtracting  (2)  from  (1),  §  587, 

log.  (1  +  a^)-  log.  (1  -  a;)=  2(^0^  +  I  + 1  + 

or,  §474,  log.l±|  =  2(^a.  +  |  +  |+...Y  (3) 

which  is  true  when  x  is  numerically  less  than  1  and  when  x  =  l. 

Let  w  be  a  positive  number  whose  natural  logarithm  is  known, 
and  let  m  be  a  greater  positive  number  whose  natural  logarithm 
is  to  be  computed.     Then,  since 

is  positive  and  less  than  1, 

m-\-n       '^ 

this  value  may  be  substituted  for  x  in  (3). 

.      m  —  n 
-|.„     _m  —  n    1+x  _        m  +  n_2m_m, 
~  m  +  n    1  —  x      .      m  —  n      2n~  n^ 
m-\-n 

and  in  (3),  since  log.  ^— ^  =  log,  —  =  log.  m  —  log.  n, 

1  1  ■  of*'*  —  "  .  IM  — nV      1/m  — wV          1     ,.. 

Iog,«  =  log,»  +  2[_^^^  +  -(^-^j  +S(^  +■■■}    W 

This  is  the  logarithmic  formula  for  m  >  n  >  0. 


EXPONENTIAL   AND  LOGARITHMIC  SERIES       529 

Since  log^  1  =  0,  by  substituting  1  for  n  and  2  for  m,  log,  2  may 
be  found ;  then  by  substituting  2  for  n  and  3  for  m,  log^  3  may  be 
found ;  etc.  Hence,  a  table  of  logarithms  may  be  constructed  by 
substituting  for  n  in  (4)  the  successive  values  1,  2,  3,  4,  •••  and 
for  m  values  greater  by  1  in  each  instance. 

Substituting  w  -f  1  for  m  in  (4)  gives  the  more  convenient 
formula, 


(5) 


log,(»+l)  =  log.n+2[^  +  g^jJ^  + g^^J;^  +  . 

which  is  true  for  all  positive  values  of  n. 

Since  the  ratio  of  convergency  of  the  series  is  always  less  than 

-,  the  series  is  very  rapidly  convergent. 

{2n-\-iy 

Thus,  when  w  =  1,  the  series  is  more  rapidly  convergent  than  the  geo- 
metrical series  ^  +  jV  +  1J5  +  ••• ;  when  n  =  2,  the  series  is  more  rapidly 
convergent  than  the  geometrical  series  ^  +  yjj  +  yy^jj  +  ••• ;  etc. 


Examples 

1.    Find  the  natural  logarithm  of  2  to  the  nearest  sixth  decimal 
place. 

Solution.  —  Substituting  1  for  n  and  0  for  logj  1  in  the  formula  for 
log.rn-l-1), 

^  V3      3.38^5.36      7.37^      ) 

_2         2  2  2 

~  3  "^  3T38  "^  6T35  "'TTs^  "*"  ■■"* 

2      2  2       2  2       2 

Since  —  =  -  h-  9,  also  —  =  — h  9,  also  —  =  — r-  9,  etc.,  and  since  these 
38     3  35     33  3'     38 

quotients  are  to  be  divided  by  1,  3,  5,  7,  •••,  respectively,  the  computation 

may  be  arranged  conveniently  as  follows  : 

3  I  2.00000000 


Adding, 


9 

.66666667  -f-  1  =  0.66666667 

9 

.07407407  --  3  =  .02469136 

9 

.0082.3045  --  5  =  .00164609 

9 

.00001449-7  =  .0001.3064 

9 

.00010101  -9  =  .00001129 

9 

.00001129  s-  11  =  .00000103 

ADV.  ALG. 3 

.00000125  -  13  =  .00000010 
loge  2  =  0.69314718 
4 

630        EXPONENTIAL  AND  LOGARITHMIC  SERIES 

The  error  committed  in  omitting  the  rest  of  the  series  is 
2  2  2 


15.3^6      17-3"      19.319 


2/11  \  2  1 

which  is  less  than  {\  A 1 h-'-liOr x  — ^—  , 

15.31H        9     92         /'        16-315     i_i' 


9 
X  -,  or 


4.    log,  5. 

6.    log,  7. 

8.    log.  9. 

5.   log,  6. 

7.    log,  8. 

9.  iog,m 

15-316     8'       15-313.8' 

2 
■which  is  much  less  than  , ,  or  .00000010,  the  last  term  found. 

Hence,  the  part  of  the  logarithm  omitted  does  not  affect  the  sixth  decimal 
place,  and  log,  2  =  0.693147  to  the  nearest  sixth  decimal  place. 
Compute  to  the  neaxest  sixth  decimal  place : 

2.  log,  3. 

3.  log,  4. 

Suggestion.  —  The  logarithm  of  a  composite  number  is  equal  to  the  sum 
of  the  logarithms  of  its  factors. 

Prove  the  following : 

10.  log.a-log,6  =  _-+^(^^+-(^__J+..., 

11.  log.Va?^  =  log.x-(^  +  jL  +  _L.+  ...^,when,r>l. 

606.  The  base  e  arises  naturally  in  the  process  of  finding  a 
formula  for  computing  logarithms.  In  this  and  in  other  theoreti- 
cal work,  natural  logarithms  are  used.  But  in  numerical  calcula- 
tions, common  or  Briggs  logarithms  are  the  most  convenient, 
because  the  base  of  the  common  system  is  the  same  as  the  base  of 
our  decimal  system  of  notation. 

Hence,  the  next  problem  is  to  discover  how  to  change  natural 
logarithms  to  common  logarithms. 

607.  To  find  the  logarithm  of  a  number  to  any  given  base. 

Let  iV  be  a  number  whose  logarithm  to  the  base  a  is  sought. 
By  §  605,  both  log,  N  and  log,  a  may  be  computed,  and,  there- 
fore, are  regarded  as  known  numbers  in  this  discussion. 

Suppose  that  log,  N=l,   or  N=  e',  (1) 

and  that  log,  a  =  k,  or  a  =  e*.  (2) 


EXPONENTIAL  AND  LOGARITHMIC   SERIES      531 

Further,  let  r  be  the  multiplier,  as  yet  unknown,  by  which 
log,  N  is  multiplied  to  produce  log„  N;  that  is,  let 

log„iNr=rlog,  xV=rZ.  (3) 

By  (3),  N=ar\ 

and  by  (2),  N=  {eY  =  e*''-  (4) 

By  (4)  and  (1),  e^'  =  e'; 

whence,  krl  =  I,  and  r  =  - ,  a  constant.  (5) 

A; 

The  constant  multiplier  by  which  the  logarithms  in  any  system 
are  derived  from  the  corresponding  natural  logarithms  is  called 
the  Modulus  of  the  given  system.  The  modulus  of  the  natural 
system  is  1. 

To  find  the  modulus  of  any  given  system,  let  a  be  the  given 
base. 

Since  by  (5),  »•  =  7^  and  by  (2),  A;  =  log,  a, 

fc 

the  modulus  is  r  = Hence, 

log.  a 

Principle  1.  —  TTie  modulus  of  any  system  of  logarithms  is  equal 
to  the  reciprocal  of  the  natural  logarithm  of  its  base. 

608.    Let  M  denote  the  modulus  of  the  common  system. 
Since  by  actual  calculation  log,  10  =  2.30258509- ••, 

by  Prin.  1,        M  =  ,-^--r  = =  .43429448.... 

^  '  log,  10     2.30258509... 

Hence,  by  the  definition  of  the  modulus  of  a  system, 

Principle  2.  —  TTie  common  logarithm  of  a  number  is  equal  to 
its  natural  logarithm  multiplied  by  .43429448.... 

Therefore,  by  (4)  and  (5),  §  605,  when  m  >  n  >  0, 

1  1  ,   o  Tirf'^n  —  71  ,  1/m  —  ri\^  ,  1/m  —  n\^  , 

logio  m  =  logio  n+2M   — ^  +  oi T"     +  S  — T"    + 

[_m  +  n     o\m  +  ny      o\m  +  nj 

and  logio  (n  + 1) 


logio n+2  M 


1       +!_,+         !__,  +  , 


_2«  +  l      3(2  71  + 1)3     5(2  n  +  1) 
in  which  Jlf=  .43429448.- 


,(1) 
,  (2) 


532 


EXPONENTIAL  AND  LOGARITHMIC  SERIES 


Examples 
1.   Compute  the  common  logarithm  of  11  to.  six  decimal  places. 
Solution.  — By  (2),  the  formula  for  the  common  logarithm  of  n  +  1, 


logioll  =logiolO  + 


(^ 


1 


21  "*"  3  •  218 


+  ■ 


1 


5.216 

For  convenience  the  computation  is  arranged  as  follows : 
logio  10         =  1.00000000 


+ 


21 


21 


21 


.86858896 


.04136138  H-  1  =    .04136138 
.00196959 

.00009379  -  3  =    .00003126 
Adding,  logio  U  =  1.04139264 

In  the  above  computation  only  the  first  two  terms  of  the  series  are  used. 

Since  the  third  term  is  less  than  —  of  the  second  term,  or  less  than  lir  of 

212  '  **i 

.00003126,  say  less  than  .00000008,  and  the  fourth  term  is  less  than  ^\j  of 
the  third  term,  and  so  on,  it  is  evident  that  the  terms  after  the  second  cannot 
affect  the  sixth  decimal  place. 

Hence,  logio  H  =  1.041393,  to  the  nearest  sixth  decimal  place. 

2.    Compute  the  common  logarithms  of  2,  3,  and  7,  and  verify 
the  following  table : 


Number 

Logarithm 

Number 

Logarithm 

Number 

Logarithm 

1 

0.000000 

5 

0.698970 

9 

0.954243 

2 

0.301030 

6 

0.778151 

10 

1.000000 

3 

0.477121 

7 

0.845098 

11 

1.041393 

4 

0.602060 

8 

0.903090 

12 

1.079181 

3.  Find  the  four-place  common  logarithms  of  the  composite 
numbers  from  14  to  25  inclusive. 

4.  Find  the  common  logarithms  of  225,  175,  and  .014  to  four 
places  of  decimals. 

5.  By   using  the   above   table    find    the    four-place    common 
logarithms  of  .125,  46.2,  1.62,  .0625,  j\,  9|,  1.1,  and  yf^. 

6.  Show  that  logi,  18  =  ^f^±-^{^. 

^^  2  log,  2  -f  log,  3 

7.  Given  log. 2  =  0.69314718  •••,  to  find  logj  7  and  log2  8. 


SUMMATION   OF    SERIES 


609.  The  summation  of  arithmetical  and  geometrical  progres- 
sions and  of  infinite  decreasing  geometrical  series  has  been  dis- 
cussed in  an  earlier  chapter.  Other  series  may  be  summed  by 
somewhat  similar  devices,  as  will  now  be  shown. 

RECURRING  SERIES 

610.  In  any  geometrical  series,  as  a  +  ax-{-  oaf  +  •••,  each  term 
after  the  first  is  equal  to  x  times  the  preceding  term.  In  the 
series  1  +  2 x—  x^  +  S jc^  —  19 x*  +  62 a^  —  •  •  •  each  term  after  the 
third  is  equal  to  the  algebraic  sum  of  —  2  a;  times  the  preceding 
term  and  3  a^  times  the  term  preceding  that.  For  example, 
8x'=-2x(-a^)+3x^(2x);    -19  x*:=-2  x(Saf)+3  x^(-a^ -,   etc. 

A  series  in  which  every  term,  after  a  certain  term,  is  equal  to 
the  algebraic  sum  of  the  products  formed  by  multiplying  the  r 
preceding  terms,  respectively,  by  r  fixed  multipliers  is  called  a 
Recurring  Series  of  the  rth  order. 

A  geometrical  series  is  a  recurring  series  of  the  first  order,  and  is  the  sim- 
plest kind  of  a  recurring  series.     The  series  l4-2x— ac^+Sx^— 19a;*-t-62a;S 

is  a  recurring  series  of  the  second  order. 

611.  If  Un  represents  any  term,  after  the  third,  of  the  series 

l  +  2x-a^  +  Sa^-19x*  +  62x^ ,     ^^ 

then,  ti„  =  —  2x  (w„_i)  +3x^  (w„^2)-         ""^^ 

Transposing,  etc.,  1  («„)  +2x (w„_i)  —  3 a^ (u^-z)  =  0.  (1) 

Equation  (1)  states  the  relation  existing  between  any  three 
terms  of  the  series,  n  being  greater  than  3 ;  and  the  expression 

1  4-  2  a;  -  3  a^, 
formed  by  taking  the  coefficients  of  w„,  u„^i,  and  m„_2  in  this  rela- 
tion, is  called  the  Scale  of  Relation  of  the  series. 

633 


534  SUMMATION   OF  SERIES 

612.   To  find  the  scale  of  relation. 

Examples 

1.  Find  the  scale  of  relation  of  the  series 

1  +  a, _|.  4a^  +  lOa^  +  22a;*  +  460?^  +  .... 

Solution.  —  Since  the  series  is  evidently  not  a  recurring  series  of  the 
first  order,  try  it  for  a  recurring  series  of  the  second  order,  and  assume 
1  +  Ax  +  Bx^  as  the  scale  of  relation. 

If  this  supposition  is  correct,  each  term,  after  a  certain  term,  increased 
by  Ax  times  the  preceding  term  and  Bx'^  times  the  term  preceding  that,  must 
be  equal  to  zero.    Therefore,  beginning  with  one  of  the  following  equations 
each  succeeding  equation  must  be  satisfied  by  the  same  values  of  A  and  B. 
4 x2  +  Ax{x)       +  B3c\l)       =0,  or    4+      A+       B  =  0 
10x^  +  Ax(ix^)   +Bz^(x)       =0,  orlO+    ^A+      B  =  0 
22  3^  +  Ax{10  x3)  +  Bx'^(i  x2)   =  0,  or  22  +  10  ^  +    4  B  =  0 
46  «8  +  ^a;(22  x^)  +  Bx%10  x^)  =0,  or  46  +  22  ^  +  10  £  =  0. 
No  values  of  A  and  B  will  satisfy  all  of  these  equations,  but  ^  =  —  3  and 
B=2  satisfy  the  last  three,  as  may  be  discovered  by  solving  any  two  of 
them  and  trying  the  values  obtained  in  the  other. 
Hence,  the  scale  of  relation  is  1  —  3  x  +  2  x^. 

2.  Find  the  scale  of  relation  of  the  series 

Solution.  — Assuming  that  the  series  is  a  recurring  series  of  the  second 

order  whose  scale  of  relation  is  1  +  Ax  +  Bx%  it  is  found  that  no  values  of  A 

and  B  make  the  equations  that  may  be  formed  on  this  supposition  consistent. 

Not  even  the  last  three  equations  thus  obtained  are  consistent.     Therefore, 

the  series  is  next  tried  for  a  recurring  series  of  the  third  order.     Assuming 

1  4-  Ax  +  Bx"^  +  Cx^  for  the  scale  of  relation,  we  obtain,  as  in  the  preceding 

example, 

^'3+     A-2B+     C  =  0,  -i-3A  +  SB+C  =  0, 

-3  +  3^+     B-2C  =  0,  7-4^-3B  +  3C  =  0. 

All  of  these,  after  the  first,  are  satisfied  by  ^  =  1,  S  =  2,  C  =  1. 

Hence,  the  scale  of  relation  is  1  +  x  +  2  x^  +  x^. 

Find  the  scale  of  relation  of 

3.  l-\-2x-\-2x^  +  2a^  +  2x*-^'". 

4.  l  +  2x  +  Sa^  +  ^a^  +  5a^-\-6T^  +  "'. 

6.  l-\-x-2x^-2c(^  +  7x*  +  7a^-20a^ . 

7.  l  +  x-\-x'-ii^~5x*-llx!'-15a^-9a^-\-—. 


SUMMATION   OF  SERIES 


535 


613.    To  find  the  sum  or  generating  function. 

By  dividing  the  numerator  by  the  denominator,  or  by  employ- 
ing the  method  of  undetermined  coefl&cients  (§  592),  certain  frac- 
tions may  be  developed  into  series  of  ascending  powers  of  a;. 

The  present  problem  is,  given  the  series,  to  find  the  generath^g 
fraction,  or  generating  function. 

Let  it  be  required  to  find  the  sum  or  generating  function  of 

l+a;  +  4a^  +  10a;3_,_22x*  +  46ar'-|-94a/'-f  .••. 
Multiplying  the  series  by  the  scale  of  relation,  1  —  3  ic  -f-  2  a^, 

14.     a;  +  4  ar^  +  10  ar^  +  22  »*  -f  46  ar'  +   94  a:«  +  ••• 
1-3  a;  +  2  x" 


1  -f  1  a;  +  4 

a^  +  10  ar*  +  22 

a^  +  46 

a^+    94 

a:«  +  ... 

-3     -3 

- 12      -  30 

-66 

-138 

_  ... 

2 

+    2+8 

+  20 

+    44 

+  ••• 

l-2x  +  3aj2+    Oaf'+    Oa;*+    Oa^+      Oa;«+... 

Since  the  series  multiplied  by  the  scale  of  relation  gives  l  —  2x 
+  3x2  +  0a^  +  0x*+ ...,  or  l-2a;  +  3a^,  if  l-2a;  +  3ar'  is 
divided  by  the  scale  of  relation,  the  series  will  be  obtained. 

That  is,  — — — r  is  the  generating  function  of  the  series. 

'  l-3a;  +  2a;2  ^  ^ 

The  generating  function  is  also  the  sum  of  the  series,  if  the 
series  is  convergent. 

Similarly,  any  recurring  series   is  shimmed  by  multiplying  the" 
series  by  its  scale  of  relation  and  indicating  that  the  result  is  to 
be  divided  by  the  scale  of  relation. 

Suppose,  for  example,  that  the  scale  of  relation  of  ao  +  aix  4-  a2X^  +  asx^ 
+  •■■  \^\  -\-  px  +  qx'^  and  that  every  term  after  the  second  is  formed  by  means 
of  the  relation 

1  (a„x")  +  i3a;(a„_ia;"-i)  +  qx^(^an-2X'^-^)  =  0,  or  a„  +  pa„_i  +  ga„_2  =  0. 
Multiplying  the  sum  of  the  first  n  terms  by  I  +  px  +  qx^, 

ao+    ai  X  +    a2  x^  +  •■•  +    a„_]  x»  1 
1    +  p    x  +    q    x^ 


ao  +    ai  I X  +    02 
paol    +pai 
+  qao 


x^  + 

+ 
+ 


+    a,,-! 

+  Pan2 

+  qon-s 


Xn-1 


+  i)an-i[x» 


-f  quo         -r  •••  -^  (/w»-3 -T  qan-21       -t- t/«n-lX"^' 

ao  +  (oi  +  pao)  X  +    0    x^  + \-  Oa;»-i  +  (pon-i  +  gan-2)x»  +  qa„-ix"+^ 


536  SUMMATION  OF  SERIES 

Hence,  the  sum  of  the  first  n  terms  is 

"         1  +  pa;  +  3x^  1  +  i)«  +  3*2 

If  the  series  is  convergent,  the  second  fraction  in  the  above  value  of  8n 
approaches  zero  as  a  limit  when  n  is  increased  without  limit,  and  the  true 
sum  is 

o  _  gp  +  («!  +  pao)x 
~     \  +  px  +  qx^ 

Examples 
Find  the  sum  or  generating  function  of 

1.  l+2.'c  +  2«2_^2ar'  +  2a;^+.... 

2.  l-\-x  +  23^  +  ^u?-\-nx*-\ . 

3.  X  -  Z:x?  +  1  x"  -  11 X*  +  41:(^ . 

4.  l-^x  +  a?  +  Za?  +  lx*-\-^l T-  +  ■-. 

5.  l+a;  +  2a;2-f-4a^-|-7a;*  +  13ic^  +  24a^  +  44a;^+.... 

6.  2~^x  +  x'  +  la?-12a*-a^  +  S2a^-^2x^ . 

614.   To  find  the  general  term. 

When  the  generating  function  may  be  separated  into  partial 
fractions  with  denominators  of  the  first  degree,  the  series  may  be 
separated  into  two  or  more  geometrical  series,  and  the  general 
term  of  the  given  series  may  be  expressed  as  the  sum  of  the  gen- 
-eral  terms  of  the  component  geometrical  series. 

Examples 
1.   Find  the  general  term  of  1  +  2  a;  +  6  a^  +  22  or*  +  86  a;*  -}-  -., 

Solution.  —  The  generating  function  is  found  to  be  TL?.? 

l-5a;  +  4a;2 
Separating  this  into  its  partial  fractions,  §  594, 


l_5x  +  4a;2      i  _  g;      l-4a;      3 


\\-x]     S\\-4txl 


Since  — —  =  1  +  a;  +  x^  +  ...  +  x' +  ••. 

1  —  X 

and  — =-^  =  1  4- 4x  +  16x2  +  ...  +  4»-x'-  + 

1  —  4x 

the  (r  +  l)th,  or  general,  term  of  the  given  series  is 

f  x--  +  I  (4'xO,  or  A  (2  +  4r)xr. 


SUMMATION   OF  SERIES  537 

Find  the  general  term  and  the  twelfth  term  of 
2.    l-2a;+7i«2-20a;3+61a;* . 

4.    l  +  5a;  +  19ar'  +  65x-3  +  211a;^  +  —. 

DIFFERENCE  SERIES 

615.  If  each  term  of  a  series  is  subtracted  from  the  following 
term,  the  successive  differences  form  another  series  called  the 
Jirst  order  of  differences  ;  if  each  term  of  the  first  order  of  dif- 
ferences is  subtracted  from  the  following  term,  a  second  order  of 
differences  is  formed ;  and  so  on. 

By  continuing  this  process  an  order  of  equal  differences  ma} 
sometimes  be  obtained. 

A  series  whose  mth  order  of  differences  is  a  series  of  equal 
numbers  is  called  a  Difference  Series  of  the  /nth  order,  m  being 
finite. 

Thus,  in  the  series  1,  8,  27,  64,  126,  216,  •••,  the  successive  orders  of  dif- 
ferences are  ■  ,  „     -,,.     r.»     ^,     ^, 

1st  order,       7,  19,  37,  61,  91,  ••• 

2d  order,     12,  18,  24,  30,  ... 
3d  order,      6,     6,     6,  ••• 

Hence,  1,  8,  27,  64,  125,  216,  •••  is  a  difference  series  of  the  third  order. 

Similarly,  7,  19,  37,  61,  91,  •••  is  a  difference  series  of  the  second  order 
whose  first  order  of  differences  is  12,  18,  24,  30,  .•• ;  and  12,  18,  24,  30,  •••  ia 
a  difference  series  of  the  first  order,  also  called  an  equidifferent  series,  or  an 
arithmetical  series. 

616.  To  find  the  nth  term  of  a  difference  series. 

Let  Gi,  a^,  eta,  •••  denote  the  terms  of  a  difference  series,  and  let 

di,  c?2,  d^,  ...  denote  t\ve  first  terms,  respectively,  of  the  first,  second, 

third,  .••  orders  of  differences. 

It  is  required  to  find  the  nth  term  of  the  series,  or  a„,  in  terms 

of  tti,  di,  ^2,  ^3,  '••,  that  is,  in  terms  of  the  first  term  of  the  series' 

and  the  first  term  of  each  order  of  differences. 
Series,  a^,  a^,  a^,  a^,  -•• 

1st  differences,  ag  — Oi,  ag  — ag,  a^  —  as,  ... 
2d  differences,  ag  —  2  ag  +  ai,  a4  —  2  a^  -f-  a2,  ••• 
3d  differences,  a4  —  3  ag  +  3  a2  —  «!,  •  •  • ;  etc. 


538  SUMMATION  OF  SERIES 

Since  di  =  a2  —  Oi, 

transposing,  02  =  01  +  ^1.  (1) 

Since       ^2  =  ag  —  2  ag  +  ctj  =  Og  —  2  (oj  +  dj)  +  Oj , 
transposing,  etc.,  ag  =  aj  +  2  dj  +  dg.  (2) 

Since  d3=a4— Sag+Saa— ai=a4— 3(ai+2di+d2)+3(ai+di)  — a^, 
transposing,  etc.,  a4  =  ttj  +  3  dj  +  3  dg  +  dg.  (3) 

If,  as  suggested  by  (1),  (2),  and  (3),  tiie  coefficients  in  the 
formula  for  the  nth  term  are  the  same  as  the  coefficients  in 
the  expansion  of  the  (n  —  l)th  power  of  a  binomial,  then,  the 
formula  for  the  nth.  term  is 

a.=a.+  (.-l)d.+  (''-V<''-^>c;,+  (''-^)C-^>(''-«).i,+  ....  (4) 

Assuming  that  this  formula  is  true  for  the  nth  term  of  a  dif- 
ference series,  the  nth  term  of  the  first  order  of  differences,  which 
is  a  difference  series  having  dg,  dg,  •••  for  the  first  terms  of  its 
orders  of  differences,  respectively,  is 

«„+!  -a„  =  d,  +  (n-  l)d2-f-^''~^^i''~^^dg  +  ....  (5) 

Adding  (4)  and  (5), 
a„+i  =  (h  +  [(n  -  1)  +  1]  d,  +  ^^[(n  -  2)  +  2]  d,, 

,      ,    ,  n(n  —  1)  ■,    ,  n(n  —  l)(n  —  2),    ,  ,„. 

=  Oi  +  wdi  +    ^  ,„     ^d2  +  -!^ -^ ^dg  +  ....  (6) 

Since  (6)  has  the  same  form  as  (4),  n  +  1  simply  taking  the 
place  of  n,  if  the  assumed  law  is  true  for  the  nth  term  it  holds 
for  the  (n  +  l)th  term.  This  law  is  true  for  the  fourth  and  pre- 
ceding terms,  as  shown  in  (1),  (2),  and  (3).  Hence,  it  holds  for 
the  fifth  term,  and  being  true  for  the  fifth  term  it  holds  for  the 
sixth,  and  so  on. 

Hence,  (4)  is  the  formula  for  any  term. 


SUMMATION  OF  SERIES  539 

617.  To  find  the  sum  of  n  terms  of  a  difference  series. 

To  find  the  sum  of  n  terms  of  the  series  Oi,  a,,  a^,  •••,  a„,  form 
an  auxiliary  series  of  which  the  given  series  is  the  first  order  of 
differences,  as 

0,  a^,  Oi  +  a2,  «i  +  ^2  +  Osj  •",  «i  +  Oa  +  ag  H h  a„. 

Then,  the  sum  of  the  first  n  terms  of  the  given  series  is  the 
same  as  the  (n  +  l)th  term  of  the  auxiliary  series,  which  may  be 
found  by  substituting  0  for  a^  a^  for  dj,  dj  for  d^,  etc.,  in  (6),  the 
formula  for  the  (n  +  l)th  term. 

Denoting  the  sum  of  the  first  n  terms  by  >S„, 

o               ,  n(n  —  V)  J    ,  n(n  —  V) (n  —  2^  ,    ,  /rr\ 

^„=wai  +  -A-^ — !-d^  +  ^ -^ '-d.,+  '-.  (7) 

Examples 

1.  Find  the  10th  term  and  the  sum  of  10  terms  of  the  series  1, 

2,  6,  15,  31,  .... 

Solution 

Series,  1,  2,  6,  15,     31,     ... 

Ist  differences,  1,  4,  9,  16,     ••• 

2d  differences,  3,  5,  7,  ••• 

3d  differences,  2,  2,  ••• 

Therefore,  by  (4),  aio  =  1  +  9(1)  + 1^  (3)  +  \^^  (2)  =  286  ; 

and  by  (7),      Sxo  =  10(1)  +  yfj  (1)  +  yfl^y  (3)  +  ^^.'^^.g^.' J  (2)  =^  836. 

Note.  —  Since  each  term  of  the  fourth  and  of  each  succeeding  order  of 
differences  is  0,  all  the  terms  after  the  fourth  of  (4)  and  (7)  vanish. 

Sum  to  12  terms  and  find  the  12th  term  : 

2.  P,  2^  3^  43,  ....  5.   2,  8,  18,  33,  54, .... 

3.  l^  2*,  3^  4^ ....  e.  2, 7, 14, 23, 34, .... 

4.    1,  8,  21,  40,  65,  ....  7.    l^  2%  3',  4%  5',  6',  7',  .... 

618.  Piles  of  spherical  shot. 

1.  If  the  base  of  a  pile  of  spherical  shot  is  a  triangle  with  n 
shot  on  a  side,  the  next  course  has  w  —  1  shot  on  a  side,  the  next 
n  —  2,  and  so  on ;  and  if  the  pile  is  complete,  the  top  course  con- 
sists of  a  single  shot.     Calling  this  the  first  course,  the  second 


640  SUMMATION  OF  SERIES 

course  has  2  shot  on  a  side  and  contains  2  +  1  shot ;  the  third 
has  3  shot  on  a  side  and  contains  3  +  2  +  1  shot ;  the  fourth 
has  4  shot  on  a  side  and  contains  4  +  3  +  2  +  1  shot ;  and,  in 
general,   the    nth    course    has   n   shot   on  a   side   and   contains 

w  +  (ri  -  1)  +  (n  -  2)  H h  2  +  1  shot,  or  |n(7i  +  1)  shot. 

Hence,  when  a  pile  of  spherical  shot  has  the  form  of  a  triangu- 
lar pyramid  the  number  of  shot  is 

1  +  3  +  6  +  10  +  15  +  -  +  i  n(7i  +  1). 

Summing  as  a  difference  series, 

series,  1,     3,     6,     10,     15,     ••• 

1st  differences,  2,     3,     4,       5,     ••• 

2d  differences,  1,     1,     1,     ••• 

.-.  §617,(7),  ^„  =  n.l+K^-^).2+^(^-^X^-^).l, 

\2  [3 

or  S,  =  in(n  +  l){n  +  2).  (1) 

2.  If  the  base  is  a  square,  the  top  shot  rests  upon  4  shot ;  these 
rest  upon  9  shot ;  and  in  general,  the  nth  course  has  n  shot  on  a 
side,  and  contains  n.^  shot. 

Hence,  when  a  pile  of  spherical  shot  has  the  form  of  a  square 
pyramid,  the  number  of  shot  is 

1  +  4  +  9 +  16 +  .-.+n2. 

Summing  as  in  1,  S„  =  ^  n(n  +  1)  (2  n  +  1).  (2) 

3.  If  the  base  is  a  rectangle  n  shot  in  width  and  m-\-n  shot  in 
length,  the  top  course  is  a  row  of  m  +  1  shot ;  the  second  course 
contains  2  (m  +  2)  shot ;  the  third,  3  (m  +  3)  shot ;  and  so  on. 

Hence,  the  number  of  shot  in  a  wedge-shaped  pile  with  a  rec- 
tangular base  is 

1  (m  +  1)  +  2(m  +  2)  +  3(m  +  3)  +  ...  +  n(m  +  n). 
Summing  as  a  difference  series, 
series,  (m+1),     2  m +  4,    3m +  9,    4m +  16,     ••• 

1st  differences,    m  +  3,         m  +  5,       m  +  7,     ••• 
2d  differences,     2,  2, 

.-.§617,   ^„  =  n(m  +  l)+^<^^(m  +  3)  +  ^^^-^)^^^g).2, 

or  /S„=^n(n  +  l)(3m  +  2n  +  l).  (3) 


SUMMATION  OF  SERIES  541 

Problems 

1.  How  many  cannon  shot  are  there  in  a  triangular  pyrami- 
dal pile  whose  bottom  course  has  10  shot  on  a  side  ? 

2.  The  base  of  a  pyramid  of  round  10-inch  shot  is  10  feet 
square.     How  many  shot  does  it  contain? 

3.  How  many  shot  are  there  in  a  wedge-shaped  pile  whose 
bottom  course  is  a  rectangle  9  shot  in  length  and  7  shot  in  width  ? 

4.  How  many  8-inch  shot  can  be  piled  on  a  rectangular  plot 
of  ground  12  feet  long  and  10  feet  wide  ? 

5.  Find  the  number  of  balls  required  to  complete  a  trian- 
gular pyramid  having  7  balls  in  each  side  of  the  top  layer. 

6.  Find  the  number  of  balls  required  to  complete  a  square 
pyramid  having  100  balls  in  the  top  layer. 

7.  Find  the  number  of  courses  of  shot  in  a  triangular  pyrami- 
dal pile  containing  165  shot. 

8 .  How  many  balls  are  there  in  an  incomplete  wedge-shaped  pile 
having  77  shot  in  the  bottom  layer  and  21  shot  in  the  upper  layer  ? 

9.  A  wedge-shaped  pile  of  shot  consisting  of  10  layers  con- 
tains 605  shot.     How  many  shot  are  in  the  top  row? 

10.  The  number  of  shot  in  a  square  pyramid  is  f^f  of  the  num- 
ber in  a  triangular  pyramid  having  the  same  number  of  shot  in 
each  side  of  the  base.     How  many  shot  are  there  in  each  pile  ? 

11.  A  fruit  seller  had  two  pyramids  of  oranges  composed  of 
the  same  number  of  layers.  One  was  a  square  pyramid  and  the 
other  a  triangular  pyramid,  and  the  square  pyramid  contained  84 
more  oranges  than  the  triangular  pyramid.  How  many  oranges 
were  there  in  each  ? 

12.  A  fruit  seller  has  a  triangular  pyramid  of  oranges,  and 
wishes  to  make  of  them  four  square  pyramids.  How  many 
oranges  must  he  use  for  the  base  of  each  of  the  square  pyramids, 
if  the  triangular  pyramid  has  2n  layers  ? 

619.  Interpolation. 

620.  The  process  of  inserting  between  two  terms  of  a  series 
numbers  that  obey  the  law  of  the  series  is  called  Interpolation. 


642 


SUMMATION  OF  SERIES 


621.  When  the  law  of  the  series  is  known,  numbers  may  be 
interpolated  between  any  two  terms  by  applying  the  law. 

Thus,  in  the  arithmetical  series  1,  9,  17,  25,  •••,  three  numbers  may  be 
inserted,  or  interpolated,  between  17  and  25  by  substituting  3|,  3^,  and  3| 
successively  for  n  in  the  formula  I  =  a +(n —  l)d  =  \ +{n  —  l)%.  The 
three  numbers  obtained,  namely,  19,  21,  and  23,  may  be  regarded  as  the 
(3J)th,  (3|)th,  and  (3|)th  terms  of  the  series,  respectively. 

When  the  law  of  the  series  is  not  known,  intermediate  terms 
may  be  found  approximately  by  treating  the  series  as  a  difference 
series,  and  giving  n  the  proper  values  in  the  formula  for  the 
nth  term. 

Examples 

1.  Given  log  20  =  1.3010,  log  21  =  1.3222,  log  22  =  1.3424, 
log  23  =  1.3617,  log  24  =  1.3802 ;  to  find  log  22.5  to  the  nearest 
ten-thousandth. 

Solution 

Series,  1.3010,  1.3222,  1.3424,  1.3617,  1.3802,  ... 

1st  difierences,  .0212,  .0202,  .0193,    .0185,  — 

2d  differences,  -  .0010,  -  .0009,  -  .0008,  ... 

3d  differences,  .0001,  .0001,  — 

Hence,  oi  =  1.3010,  di  =  .0212,  c?2  =-  .001,  and  dz  =  .0001  ;  also,  since 
22.5  lies  halfway  between  22  and  23,  whose  logarithms  are  the  third  and 
fourth  terms  of  the  series,  n  =  3.5. 

Substituting  these  values  in  the  formula  for  the  nth  term,  §  616,  (4), 

log  22.5  =  1.3010  4-  .053  -  .001875  +  .00003126  =  1.3522. 

Note.  —  In  the  above  example  the  third  differences  are  not  absolutely 
equal  to  each  other,  as  may  be  seen  by  taking  the  given  logarithms  to  more 
than  four  places  of  decimals,  and,  therefore,  the  fourth  differences  are  not 
equal  to  zero.  But  the  fourth  differences  and  all  succeeding  them  are  so 
small  that  they  may  be  disregarded  when  log  22.5  is  required  to  only  four 
decimal  places. 

2.  In  the  following  table  the  surfaces  and  volumes  of  spheres 
whose  diameters  are  1,  2,  •.-,  5  are  given.  Find  the  surfaces  and 
volumes  of  the  spheres  whose  diameters  are  3.1,  3.2,  3.7. 


Diam. 

1 

2 

3 

4      5 

Surf. 

3.1416 

12.5664 

28.2744 

50.2656  78.5400 

Vol. 

.5236 

4.1888 

14.1372 

33.5104  65.4500 

SUMMATION  OF  SERIES  543 

3.  Given  ^350  =  7.0472987,  ^/35i  =  7.0540041,  ^/352  = 
7.0606967,  ^353  =  7.0673767,  ^354  =  7.0740440,  to  find 
the  cube  root  of  351.6. 

4.  Given  7!-^=. 00 1379310,  yi^=. 001369863,  ,4^=. 001360544, 
7-^^=  .001351351,  y^  =  .001342282,  to  find  the  reciprocal  of  736. 

SERIES  WHOSE  ^TH  TERMS  ARE  FUNCTIONS  OF  N 

622.  The  sum  of  the  first  n  terms  of  a  series  is  represented,  in 
general,  by  S^-  But  if  the  nth  term  of  the  series  is  a  function  of 
n,  the  sum  of  the  first  n  terms  may  be  represented  by  prefixing 
Sn  to  the  nth  term. 

Thus,  SnU^  means  the  sum  of  the  first  n  squares;  65(211 +  1)  means. 
14-3  +  5  +  7  +  9,  the  sum  of  the  first  5  odd  numbers. 

623.  Summation  by  undetermined  coefficients. 

When  the  nth  term  of  a  series  has  the  form  a +  1)71  +  en-  +  •••, 
the  series  may  be  summed  to  n  terms  by  employing  undetermined 
coefficients. 

Examples 

1.    Find  the  sum  of  the  first  n  squares. 

Solution.  — Since  each  term  of  1^  +  2^  +  3^  +  ...  +  rfi  is  a  function  of  n, 
assume  S„.n'^  =^  A+  Bn  +  Cn^  +  Dn^  +  En*^  +  •••  (1) 

and    ^„+i?i2  =  ^  +  5(?i  + 1)  +  C(n  + 1)2  + i)(n+ 1)3  +  ^(74  + 1)4  +  ....    (2) 

Subtracting  (1)  from  (2), 
(n  +  l)2=J5+C(2n  +  l)+Z)(3n2+3„+i)^.^(4,i34.6„2+4„  +  i)  +  ..._      ^3^ 

Since  the  first  member  involves  no  power  of  n  higher  than  the  second,  E 
and  all  succeeding  undetermined  coefficients  vanish,  and 

n2  +  2  n  +  1  =  ^  +  (7(2  n  +  1)  +  Z)(3  n2  +  3  n  +  1) 
=  (B  +  a  +  i))  +  (2  C  +  3  D) M  +  3  Z)n2  . 

.-.§590,  B  +  C  +  D  =  l,  2C  +  3Z>  =  2,  and  3D  =  1. 

Solving,  D  =  \,  C=i,  B  =  \. 

.:  12  +  22  J.32  +  ... +  n2  =  ^+  ^n  + Jn2  +  |n8 

=  ^  +  iM(n  +  l)(2n  +  l). 
Since  Sxn^  =  1,  I2  =  ^  +  J  .  1  .  2  •  3,  whence  ^  =  0. 

.-.  12  +  22  +  32+ ...  +  n2  =  ^n(n  + l)(2n  +  l). 


644  SUMMATION  OF  SERIES 

Show  by  employing  undetermined  coefl&cients.that: 

2.  1  +  2  +  3 -}-••. -fw=iri(«  + 1). 

3.  l«  +  23  +  33  +  ...  +  w3  =  ^w2(«  +  l)2. 

4.  1^  +  2^  +  3*  +  ...  +  n*  =  ^\n(n  -\-l){2n  +  l)(3n'  +  3n-  1). 

5.  1*  +  3-  +  5-  +  . . .  +  (2 n  -  1)2  =  ^  w  (2  n  +  1)  (2  w  -  1). 

624.    Compound  series. 

Some  series  may  be  summed  readily  by  separating  each  term 
Into  two  terms  and  adding  the  component  series  thus  obtained. 

Examples 
1.    Find  the  sum  to  n  terms  and  to  infinity  of  the  series 
2  2  2 

1.22.33.4 

Solution.  —  Separating  the  general  term  into  two  fractions  having  tne 
same  numerators  as  the  terms  of  the  series, 

2         ^  2  _  _2 

n(n  +  1)      n      n  -\-\ 

Since  this  is  true  for  each  term  of  the  series, 

-■=(!-i)-(i-i)-(i-l)— (!-;r^) 

2     3  n  —  \      n 

2     2  2         2         2 


=  2- 


2     3  n-\      n     n+l 

2  2n 


n+ 1      n  +  1 
As  n  increases  without  limit,  this  sum  approaches  2  as  a  limit. 

Find  the  sum  to  n  terms  and  to  infinity  of 

2.  J_  +  J_  +  J-  +  ....  5.    J-+J-  +  J-  + 

1.3  2.4^3.5  3.5      5.7      7.9 

3.  _^+-^+ -^  +  ....  6.    -?-  +  ^  +  -?-  + 

1.4  2.5      3.6  2.6     3.6     4.7 

1.5^3.7^5.9  1.3     2.4     3.5 


FUNCTIONS    OF   A    SINGLE   VARIABLE 


625.  The  general  object  of  this  chapter  is  the  comparison  of 
the  corresponding  values  of  x  and  certain  functions  of  x.  For  this 
purpose  X  is  regarded  as  a  variable  increasing  continuously  from 
an  indefinitely  large  negative  value,  denoted  by  —  go,  to  an  in- 
definitely large  positive  value,  denoted  by  +  co. 

TABULATION  OP  FUNCTIONS 

626.  By  giving  the  independent  variable  x  any  values  we 
choose,  the  corresponding  values  of  any  given  function  of  x,  as 
i(a^  —  6  x),  may  be  obtained.  The  manner  in  which  the  function 
changes  with  the  independent  variable  may  then  be  exhibited  in 
a  table  of  correspondiiig  values.     For  example. 


when     X  = 

-2 

-1 

0 

1 

2 

3 

4 

5 

6 

7      8 

\(x'-Qx)  = 

8 

31 

0 

-^ 

-4 

—  4.1 

-4 

-^ 

0 

^    8 

Since  a;^  —  6 x  =  x{x  —  6),  if  x> 6  and  increases  without  limit,  \{x'^—Q  x) 
will  be  positive  and  increase  without  limit.  This  is  denoted  by  annexing  to 
the  above  table  the  corresponding  values  a;  =  +  oo,  ^{x^  —  6  x)  =  +  oo.  Again, 
if  X  is  negative  and  decreavses  without  limit,  |(x2  —  6  x)  will  be  positive  and 
increase  without  limit.  This  is  denoted  by  prefixing  to  the  table  the  cor- 
responding values  X  =  —  00,  ^(x^  —  6  x)  =  -f-  oo. 


627.  The  symbol  for  any  given  function  of  x  is  f(x),  read 
'  function  of  .r.'  Other  functions  of  a;  in  a  discussion  may  be 
represented,  if  desired,  by  F(x),  <f>(x),  f'{x),  etc.,  read  '  large  F 
function  of  x,'  '  phi  function  of  a;,'  '/-prime  function  of  a?,'  etc. 

Since  f{x)  is  a  variable,  depending  upon  the  value  assigned  to 
X,  fix)  is  often  represented  by  the  variable  y. 


ADV.  ALG.  — 36 


545 


546  FUNCTIONS   OF  A    SINGLE    VARIABLE 

Values  of  f(x)  corresponding  to  particular  values  assumed  for 
X,  as  3,  0,  —  1,  a,  are  usually  indicated  thus :  /(3),  /(O),  /(—  1), 

If/Cx)  =  x2  -  6  x+  0,  /(0)=  9  ;  /(1)=  1  -6  +  9  =  4  ,  /(2)=4-12  +  9  =  l ; 
/(3)  =  9  -  18  +  9  =  0  ;  and  in  general, /(a)  =  a^  -  6  a  +  9. 

628.  A  convenient  method  of  substitution. 

Let  it  be  required  to  find  the  value  of  a^-\-  Aa^  +  7 x  +  G  when 
x  =  2.  The  process  of  substituting  2  for  x  is  similar  to  the 
arithmetical  process  of  reducing  a  compound  denominate  number 
to  a  number  of  lower  denominatioii. 

Process  Explanation.  —  Since  a;  =  2,  x^  =  2  a;2. 

Writing  2  x^  under  4  x-  and  adding,  the 

ar4-4ar-(-     t  x  -\-    6  |  2  sum  of  the  first  two  terms  of  /(x)  is  equal 

2  a^  +  12  .r  4-  .'58  to  6  a;2.     Since  x  =  2,  Gx"^  =  12x.    Writ- 

ar'  -I-  6  ar  4-  19  r   I   44  —  ^2^      ^^^  ^^  ^  under  7  x  and  adding,  the  sum  of 

the  first  three  terms  of  /(x)   is  equal  to 
19  X.     Since  x  =  2,  19  x  =  38.    Adding  38  to  6,  the  remaining  term  of  /(x), 
44  is  obtained  for  the  value  of /(x)  when  x  =  2  ;  that  is,  /(2)  =  44. 
In  practice  only  the  detached  coefficients  are  used,  thus  : 

1  +  4+    7+    6(_2 

2  +  12  +  38 
1  +  6  +  19  I  44     .-. /(2)=44. 

Examples 
Tabulate  the  following  functions  of  x  for  integral  values  of  x 
between  —  5  and  5  and  for  a;  =  ±  co  : 

1.    2  a;  —  1.  2.    ar'  —  a^  +  a;  -  1.  3.    .r'  -  6  a;^  +  11  x  —  6. 

GRAPHICAL  REPRESENTATION  OF  FUNCTIONS 

629.  Heretofore,  functions  of  x  have  been  given  a  purely- 
numerical  interpretation.  By  representing  the  values  of  f(x)  by- 
distances,  it  is  possible  to  give  a  pictorial  or  graphical  representa- 
tion of  the  function  passing  through  all  its  values  as  x  increases 
continuously  from  —  cc  to  +  oo.  The  function  ^(x?  —  6x)  will  be 
used  for  purposes  of  illustration. 

630.  Since  the  values  of  x  and  ^(a^  —  6  a;)  tabulated  in  §  626  are 
real,  they  may  be  represented  graphically  by  distances  from  a  zero 


FUNCTIONS   OF  A    SINGLE    VARIABLE 


547 


point  on  a  horizontal  line,  positive  distances  being  laid  off  toward 
the  right  and  negative  distances  toward  the  left.  But  to  avoid 
confusing  the  values  of  x  and  f(x),  it  is  customary  to  lay  off  the 
values  of  the  function  on  or  parallel  to  a  second  line  crossing  the 
first  at  right  angles  at  the  zero  point,  positive  values  of /(a;)  being 
represented  by  distances  measured  upward  and  negative  values 
by  distances  measured  downward  from  the  first  line. 

The  first  line  is  called  the  JIT-axis,  and  the  second  the  K-axis, 
the  function  of  x  being  denoted  by  y. 

Fig.  1  is  a  graphical  represen- 
tation of  the  manner  in  which 
the  function  \(3r  —  6  x)  changes 
with  the  independent  variable  x. 
See  the  table  of  corresponding 
values,  §  626. 

The  point  Pj,  which  is  2  units 
to  the  left  of  the  zero  point  and 
at  the  same  time  8  units  above 
it,  represents  the  state  of  the 
variable  and  the  function  when 
x  =  —  2  and  f(x)  =  8.  Similarly, 
P.2  represents  the  next  state  of 
the  variable  and  the  function,  Pg 
the  next,  and  so  on. 

The  lengths   of  the   vertical 
dotted  lines  represent  the  values  of  y,  or  f(x).     They  are  called 
ordinates.     For  example,  y^  is  the  ordinate  of  Pj  and  is  equal  to 
/(-2),or8. 

The  corresponding  horizontal  distances  are  called  abscissas. 
For  example,  —  2  is  the  abscissa  of  Pj. 

The  abscissa  and  ordinate  of  a  point  are  called  its  coordinates. 

The  point  whose  abscissa  is  a  and  ordinate  h  is  denoted  by  the 
symbol  (a,  h).     The  abscissa  is  always  written  first. 

Thus,  the  point  Pi  is  represented  by  (—2,  8). 

If  the  successive  values  of  x  between  any  two  finite  values  are 
taken  sufficiently  near  together,  the  corresponding  values  of  y 
—  i(^  —  6  a;)  may  be  made  to  differ  from  each  other  as  little  as 
we  please ;  that  is,  y  varies  continuously  with  x,  or  is  a  continuous 


Fig.  1. 


548 


FUNCTIONS   OF  A    SINGLE    VARIABLE 


X 


FUNCTIONS   OF  A    SINGLE    VARIABLE  549 

function  of  x  (§  435).  Hence,  the  assemblage  of  all  the  points  rep- 
resenting corresponding  values  of  x  and  y  form  a  continuous  line. 
This  line,  the  coordinates  of  every  point  of  which  satisfy 
the  equation  y  =  ^(af  —  6  x),  is  called  the  Locus  of  the  equation 
y  =  i(a^  _  6  X),  or  the  Graph  of  the  function  ^(x^  —  6x).  Within 
the  limits  of  the  drawing  it  is  represented  by  the  curved  line 

The  graphs  of  several  functions  of  x  are  shown  in  Fig.  2. 

631.   Roots  of  /(jr)  =  0. 

Consider  the  functions  of  x  plotted  in  Fig.  2. 

1.  Since  whenever  the  graph  of  a  function  crosses  or  touches 
the  axis  of  x  the  ordinate  of  that  point  is  equal  to  zero,  the  func- 
tion is  equal  to  zero  for  the  value  of  x  denoted  by  the  correspond- 
ing abscissa.  For  example,  a^  -1-  9  a^  +  23  a;  -f- 15  is  equal  to  zero 
when  a;  =  —  5  or  —  3  or  —  1,  as  shown  by  its  graph  crossing  the 
axis  of  X  at  these  points;  that  is,  the  graph  of  the  function 
a^-f9a:^-f-23a/'  +  15  shows  that  —5,-3  and  —  1  are  real  roots 
of  the  equation  ar*  +  9  a^  +  23  x  +  15  =  0. 

Again,  a^  —  8a;  +  16=0  when  x  =  4,  as  shown  by  its  graph 
touching  the  axis  of  x  at  this  point ;  that  is,  4  is  a  real  root  of 
the  equation  a^  —  8  a;  +  16  =  0. 

Hence,  the  real  roots  of  f{x)  =  0  are  represented  graphically  by 
the  abscissas  of  the  points  where  the  graph  of  f{x)  crosses  or  touches 
the  axis  of  x. 

2.  Since,  for  all  values  of  x,  a^  —  8  a;  +  20  is  4  greater  than 
x^  —  8  a;  -f  16  and  the  latter  is  4  greater  than  a^  —  8  a;  +  12,  the 
graphs  of  these  functions  are  exactly  alike  except  in  position. 

Suppose,  then,  that  the  graph  of  a^  —  8  a;  +  12,  which  exhibits 
two  unequal  real  roots,  2  and  6,  is  moved  vertically  to  coincide 
with  the  graph  of  .r^  —  8  a;  -f- 16.  It  is  evident  that  the  two  un- 
equal roots  become  more  and  more  nearly  equal  as  the  graph  is 
moved  upward,  and  become  equal  when  the  graphs  coincide. 

Hence,  if  the  graph  off(x)  touches  the  axis  of  x  but  does  not  cross 
it  or  terminate  in  it,  the  abscissa  of  the  point  of  tangency  represents 
two  equal  real  roots  off(x)  =  0. 

3.  Now  suppose  that  the  graph  is  moved  vertically  once  more 
toward  the  graph  of  a;^  —  8  a;  +  20.     The  graph  ceases  to  touch 


550  FUNCTIONS   OF  A    SINGLE    VARIABLE 

the  axis  of  x,  and  we  infer  that  there  is  no  real  value  of  x  which, 
substituted  for  x,  can  reduce  the  function  to  zero. 

This  inference  accords  with  the  fact  that  the  roots  of  a^  —  8  a; 
+  20  =  0  are  imaginary  (§  304,  Prin.  1). 

Hence,  if  the  graph  of  f(x)  has  no  point  lying  in  the  axis  of  x,  all 
the  roots  of  f(x)  =  0  are  imaginary. 

Examples 

Plot  the  following  functions  of  x,  and  discover  from  the  graphs 
as  much  as  possible  concerning  the  roots  of  f(x)  =  0  : 

In  the  following  examples  the  student  should  substitute  values  of  x  close 
enough  to  each  other  to  determine  the  form  of  the  graph  with  some  accuracy. 

Cross-section  paper  is  very  useful  in  this  work. 

It  is  often  convenient  to  plot  the  values  of  f(x)  to  a  smaller  scale  than 
that  employed  for  the  values  of  x. 

1.  2x  +  l.  4.    1-a^.  7.    a^-ex'  +  llx-G. 

2.  xF  —  5x  +  4:.         6.    a^-2x~l.         8.    »"  —  7 a;  +  6. 

3.  a^_6a;-l-9.        6.    x--^4.  9.    a-^ - 9 a^  +  22 ar' - 32. 

DERIVED   FUNCTIONS 

632.  Another  method  of  comparing  the  variation  of  f(x)  with 
that  of  X  consists  in  determining  for  what  values  of  x,  if  any, 
f(x)  increases  or  decreases  when  x  increases  continuously,  and 
in  finding  how  fast  f(x)  is  changing  as  compared  with  the  corre- 
sponding changes  in  x. 

633.  Let  f(x)  denote  any  function  of  x,  and  suppose  that  x  is 
increased  by  a  positive  number  h  from  the  value  a  to  the  value 
a-{-  h.     Then  h  is  called  an  increment  of  the  variable. 

Since  the  increment  given  to  x  produces  a  change  in  the  value 
of  the  function  from  /(a)  to  /(a  +  h),  the  change  in  the  function, 
f(a  +  h)  —f(a),  is  called  an  increment  of  the  function. 

In  general, /(a;  +  ^)  —f(x)  represents  any  increment  of /(x). 

When  f(x)  is  increasing,  the  successive  increments  of  the  func- 
tion are  positive,  and  when  f{x)  is  decreasing,  they  are  negative. 


FUNCTIONS   OF  A    SINGLE    VARIABLE  551 

634.  When  equal  increments  of  x  produce  equal  increments  in 
f{x),  as  in  the  case  of  the  functions  2  x  and  2  a;  +  8,  Fig.  2,  the 
rate  of  change  of  f{x)  is  uniform  and  may  be  obtained  by  finding 
the  ratio  of  any  increment  of  fix),  however  large  or  small,  to  the 
corresponding  increment  of  x.  When  f{x)  does  not  vary  uni- 
formly with  X,  the  rate  of  change  of  f{x)  with  respect  to  x  at  any 
instant  during  a  small  interval  is  obtained  approximately  on  the 
supposition  that  fix)  varies  uniformly  with  x  during  that  interval. 

The  smaller  the  interval,  then,  or  the  smaller  the  value  of  h, 

the  more  nearly  will  the  ratio  — ^ —  represent  the  true 

rate  of  change  of  f(x)  as  x  changes  from  a  to  a  +  ^. 

Hence,  the  limit  of  this  ratio  as  /i  =  0  represents  the  rate  of 
change  of  f(x)  at  the  instant  when  x  =  a. 

The  same  is  true  for  any  value  of  x.     Hence, 

-f(x  +  h)~f(xy 


lim. 


h 


h=0 


represents  the  rate  of  change  of  f(x)  at  any  instant  as  compared 
with  that  of  x. 

635.  The  limit  of  the  ratio  of  the  increment  of  f{x)  to  the  cor- 
responding increment  of  x  as  the  increment  of  x  approaches  the 
limit  zero  is  called  the  Derivative  of  the  function  with  respect  to  jr, 
or  the  First  Derived  Function  of  jr. 

The  formula  for  the  derivative  of  f(x)  with  respect  to  x  is 

■f(x  +  h)-f(xy 


lim. 


h 


n 

Jh=0. 


This  is  represented  by  —  f{x). 
ax 

Thus,  —  (x^  +  2  x)  means  the  derivative  with  respect  to  a;  of  x^  -^  2  x. 
dx  ,        n 

Other  notations  are  Dif(x),  ^-  w,  -^,  etc. 
•'^  ^'  dx      dx 

636.  Since  the  derivative  with  respect  to  x  of  f(x)  is  usually 
another  function  of  x,  which  in  turn  may  have  a  derivative  with 
respect  to  x,  which  may  have  still  another,  and  so  on,  the  first 
derivative  will  be  denoted  by/'(x),  the  second  hj  f"(x),  the  third 
by  f'"(x),  and  so  on. 

Successive  derivatives  are  sometimes  denoted  by  /i(x),  /'(x),  fz{x),  etc. 


652  FUNCTIONS   OF  A    SINGLE    VARIABLE 

These  derived  functions  are  called  the  first  derived  function  of  x, 
or  the  derivative  of  f{x)  with  respect  to  x,  the  second  derived 
function  of  x,  and  so  on. 

Examples 

1.  Find  the  derived  functions  of  a^  —  8  a;  + 12. 
Solution.  —  Let  x^  -  8  a;  +  12  =  fix) . 

Then,  f(x+ h)-fix)=  x'^+2  hx  +  h^- (8  x+8h) +  12- (x'^-8x+12) 

=  2hx-8h  +  h\ 

...   fct^^-^  =  2x-8  +  A.    f\\\\ 

TaMng  the  limit  as  /i  =  0, 

f'{x)  =  2x  —  8,  the  Jirst  derived  function. 
Again,      /'(x  + /i)-/'(x)=  [2(x  +  A)- 8]-(2x  -  8) 
=  2h. 
.    /'(x  +  /0-/'(x)_^ 
••  h  -^• 

Taking  the  limit  as  A  =  0, 

/"(x)=  2,  the  second  derived  function. 

Since  the  second  derived  function  is  a  constant,  its  increment,  and  there- 
fore its  derivative,  is  zero.  Hence,  all  derivatives  of  x^  —  8  x  -f  12  after 
the  second  vanish. 

Find  the  successive  derived  functions  of 

2.  2x.  ^.    x-  —  8x.  6.    3  —  2x  —  x^. 

3.  2x4-1.  5.    a.-2-8a;  +  16.  7.    a? +  9x^ +  23x-{- 15. 

637.  A  function  that  involves  only  integral  powers  of  x  and  is 
integral  with  respect  to  x  is  called  a  rational  integral  function  of  x. 

3  1 

x^  +  1  +  x"2  is  rational  but  not  integral  with  respect  to  x  ;  x^  +  |  x^  is 
integral  but  not  rational  with  respect  to  x  ;  x^  -f  5  x  -t-  6,  x^  -}-  i  x^  -t-  3  x,  and 
a;2  -I-  ■v/2  X  +  1  are  both  rational  and  integral  with  respect  to  x,  that  is,  are 
rational  integral  fimctions  of  x. 

638.  The  general  form  of  a  rational  integral  function  of  x  is 

aaf  +  6x"-'  +  ca;»-2  -| [-kx  +  l, 

in  which  w  is  a  positive  integer,  and  a,  b,  c,   •••,  k,  I  do  not 
involve  x.     But  it  is  more  convenient  to  use  the  form 

af  -\-px^-^  +  9cc»-2  -\ \-sx  +  t, 

which  may  be  obtained  by  dividing  the  general  form  by  a. 


FUNCTIONS   OF  A    SINGLE    VARIABLE  553 

639.   Derivative  of  a  rational  integral  function  of  jr. 

Let  f{x)  —  X"  +  2Jcc"-i  +  qx"--  H \-sx  +  t  (1) 

be  a  rational  integral  function  of  x. 
Give  X  an  increment  h.     Then, 

f(x+h)  =  (x+h)'^+pix-\rhY-'+q(x+hy-'+  •••  +s(x+h)+t.   (2) 
Expanding  and  arranging  to  ascending  powers  of  h, 

f(x+h)=  x"    +  px''-'^+  qx^-^-l \-sx-^t] 

-{-h\nx''-'^  +  (n—l)px"-^+{n—2)qx"-^-{ \-sl 


^^ln(n~l)x^-^+(n-l)(n-2)px''-^+-l 

+  ^"|n(n-l)(n-2)...l| 

\n 


■  G^) 


Subtracting  f(x)  from  the  first  member  and  its  value  given  in 
(1)  from  the  second,  dividing  the  result  by  h,  and  taking  the  limit 
of  the  result  as  /i  =  0,  the  terms  in  (3)  involving  h^,  h^,  •■-,  h" 
vanish,  and  we  have 

lim.[M=^ 


=nx^-'^-{-(7i—l)px''-^-j-(n—2)qx''-^-] \-s; 

^=0 


that  is,      —  (xf"     +  »a;"*^  +  ga;"-^  -i h  sx  +  0 

dx  ' 

=  naf-^  +  (w  —  \)px''-^  +  (w  —  2)  qx""-^  -\ \- s.  (4) 

Hence,  the  derivative  of  a  rational   integral  function  of  x  is 

obtained  by  midtiplying  each  term  of  f(x)  by  the  exponent  of  x 

in  that  term  and  diminishing  the  exponent  of  x  by  unity. 

Thus,  —  (x8  +  5  a;2  +  7  3c  _  4)  =  3  a;2  +  10  a;  +  7. 

dx 

640.    Denoting  the  successive  derived  functions  of  x  by  fix), 

f'V),  -', 

f(x)     =  ic"  -{-  p«"-^  +  qxT-'^  -\ \-sx  +  t, 

f'{x)    —  na;"-^  +  (n  —  l)pa;"  ^  +  (n  —  2)  qx''-^  -\ \-s, 

f"(x)  =n(n-  1) iC»-2  +  (w  _  1)  (n  _  2)px''-^  -j , 

f"<(x)  =n(n-l){n-2)  x^'^  +  •  •  •, 

/"(x)  =w(n-l)(n-2)...2.1.ir«  =  |w. 


654  FUNCTIONS   OF  A    SINGLE    VARIABLE 

Hence,  (3)  may  be  written 
f{x  +  h)  =f(x)  +  hf'(x)  +  f^f"(x)  +^f'\x)  +  -.  +  h\     (5) 

Formula  (5)  gives  a  process  of  substituting  x  +  h  for  x  in  a 
rational  integral  function  of  x.  It  is  a  special  case  of  Taylor's 
Formula. 

641.  Continuity  of  a  rational  integral  function  of  x. 

By  (5),  §  640,  if  f{x)  is  a  rational  integral  function  of  x, 

fix  +  h)  -f{x)  =  hf'ix)  +  |/"(a;)  +^r'{x)  +  ■••  +  h\ 

The  first  member  is  the  increment  of  f{x)  produced  by  giving 
x  the  increment  h.  To  determine  whether  the  given  function  is 
continuous,  it  is  only  necessary  to  find  whether  an  infinitesimal 
change  in  the  variable  produces  an  infinitesimal  change  in  the 
function.     Then,  let  h  =  0. 

Since  the  second  member  consists  of  a  finite  number  of  terms 
and  the  coefficient  of  each  power  of  h  is  finite  for  all  values  of  x, 
as  /i  =  0  the  second  member  approaches  zero  as  a  limit. 

Hence,  every  rational  integral  function  of  x  is  continuous  for  all 
finite  values  of  x. 

642.  Derivative  of  the  product  of  two  or  more  functions. 

Let  y  and  z  be  rational  integral  functions  of  a;  and  let  y^  and 
z'  be  particular  values  of  y  and  z  corresponding  to  a;  =  a. 

To  find  the  derivative  of  yz  with  respect  to  x,  suppose  that,  as 
X  increases  from  a  to  o  +  h,y  changes  from  y'  to  y'  +  A;  and  z  from 
z'  to  z'  +  I     If  /i  =  0,  then,  §  641,  A;  =  0  and  Z  =  0. 


When  x  =  a,  — yz  =  lim. 
dx 


\y'  +  k){z'  +  l) 


^1 

J/t  =  0 


simplifying,  =  lim. 


h 
■kz'  +  (y'  +  k)r 
h 
Then,  since  y'  and  z'  are  constants,  and  as  h  =  0,  y'  -\-  k  =  y', 

h 


—  yz  =  z'  Jim. 
dx' 


h  =  0 


'1 

Jh  =  0 
+  y'lim.    - 


§635,  =z'^y  +  y'^z. 

dx  dx 


i 


FUNCTIONS   OF  A    SINGLE    VARIABLE  565 

Since  this  is  true  for  each  particular  value  of  x, 

d  cl      ,      d 

—yz  =  z--y  +  y—z. 
dx  dx  dx 

Similarly,   —  v(yz)  =yz  —  v  -\-v — yz 
dx  dx  dx 

d      ,        d      ,        d 
=  yz—v  +  zv—y  +  vy—z, 
dx  dx  dx 

and  so  on  for  any  number  of  variable  factors.     Hence, 

The  derivative  toith  respect  to  x  of  the  product  of  any  number  of 

rational  integral  functions  of  x  is  equal  to  the  sum  of  the  products 

obtained  by  multiplying  the  derivative  of  ea/ih  function  by  the  product 

of  all  the  other  functions. 

Thus,  —  (x  -  a)(x  -  b)(x  -c)  =  (x-  6)(x  -  c)—  (x  -  a) 
dx  dx 

+  (x  -  c)  (x  -  «)  —  (x  -  6) 
dx 

+  (x-a)(x-b)  —  (x-  c) 
dx 

=  (x  —  6)  (x  —  c)  +  (x  —  c)  (x  —  a) 
+  (x  —  a)  (x  —  b). 

Examples 

1.  Find  the  derivative  with  respect  to  x  of  (x—l)(x—2)(x—3). 
Test  the  result  by  comparing  it  with  the  derivative  with  respect 
to  X  of  the  expanded  product. 

2.  Find  the  first  and  second  derived  functions  of  (x  —  l)^(a;  +  2) 
and  express  each  in  the  form  (x  —  l)"^(x). 


MAXIMA  AND  MINIMA 

643.  In  tracing  the  graph  of  3  —  2x—  a^,  Fig.  2,  conceive  x  to 
increase  continuously  from  —  oo  to  +qo. 

As  X  increases  continuously  up  to  —1,  the  function  3  —  2  a;  —  o;^ 
increases  continuously  up  to  4,  and  as  x  increases  continuously 
from  —1  to  -f  CO,  the  function  decreases  continously  from  4  to 
—  CO.  Hence,  3  —  2  a;  —  x^  is  said  to  be  an  increasing  function  of  x 
for  values  of  x  less  than  —1  and  a  decreasing  function  of  x  for 
values  of  x  greater  than  —1. 


656  FUNCTIONS   OF  A    SINGLE    VARIABLE 

.644.  Any  value  at  which  a  continuous  function  of  a  variable 
ceases  to  be  an  increasing  function  and  begins  to  be  a  decreasing 
function  is  called  a  maximum  of  the  function.  For  example,  4  is 
a  maximum  oi  3  —  2  x  —  x^. 

Any  value  at  which  a  continuous  function  of  a  variable  ceases 
to  be  a  decreasing  function  and  begins  to  be  an  increasing  function 
is  called  a  minimum  of  the  function.  For  example,  —  4  is  a 
minimum  of  x*^  —  8  a;  +  12. 

The  maxima  and  minima  of  a  function  are  called  its  turning 
values,  or  critical  values. 

Cautiox.  —  A  maximum  of  a  function  is  not  necessarily  its  gi-eatest  value 
nor  a  minimum  its  least  value,  as  may  be  seen  in  the  graph  of  x^  +  9  x^  +  23  a; 
+  15.     (Fig.  2.) 

645.  The  test  for  turning  values  is  as  follows : 

If  f(a)  is  greater  than  either  f(a  +  h)  or  f{a  —  h)  however  small 
h  may  be  taken,  then  f(a)  is  a  maximiim  of  f{x). 

If  f(a)  is  less  than  either  f(a  +  h)  or  f(a  —  h)  however  small  h 
may  be  taken,  then  f(a)  is  a  minimum  of  f(x). 

Thus,  3  —  2  a;  —  a;2  is  greater  for  x  =  —  1  than  for  any  other  values  of  x  in 
the  vicinity  of  —  1,  however  little  they  may  differ  from  —  1.  Hence,  /(—  1), 
or  4,  is  a  maximum  of  3  —  2  x  —  x^. 

646.  Let  ax^  +  bx  -{-  c  be  any  quadratic  function  of  x  having 
real  coefficients.  Placing  the  function  equal  to  y  and  solving 
for  X, 

b    ,    1 


a;  =  - -^  ± -^  V4  ay  -  (4  oc  -  6=^) . 

2  a     2  a 

Considering  only  real  values  of  x,  4  ai/  —  (4  ac  —  6^)  must  be 

positive  or  else  equal  to  zero  ;  that  is, 

4  ay  -  (4  ac- 62)^0. 

Therefore,  §§  401,  403, 

if  a  is  positive,  y>:c , 

~        4a 

b^ 
but  if  a  is  negative,  2/  ^  c  —  - —  • 

In  the  first  case  y,  or  ax^  +  bx  -\-  c,  may  be  as  great  as  we 

please,  but  cannot   be  less  than  c  —  -— ,  which   is  therefore  a 

.   .  4a 

minimum. 


FUNCTIONS   OF  A    SINGLE    VARIABLE  557 

In  the  second  case  the  function  may  be  as  small  as  we  please, 

but  cannot  be  greater  than  c ,  which  is  therefore  a  maximum. 

Hence,  ^  ^ 

Principle  1.  —  Every  quadratic  function  of  the  form  aa?  -\-hx-\-c 
has  one  and  only  one  critical  value,  c  —  (6^  -h  4  a),  which  is  a  mini- 
mum or  a  maximum  according  as  a  is  positive  or  negative.  TJie 
coiTesponding  value  of  x  is  —h  -i-2  a. 

Thus,  x^  —  8  X  +  12  has  only  one  critical  value,  12  —  (64  ^4),  or  —  4, 
and  this  is  a  minimum  because  the  coefficient  of  x^  is  positive  ;  the  corre- 
sponding value  of  X  is  8  -T-  2,  or  4.  Again,  3— 2x— x^  has  only  one  critical 
value,  3—  (4-: — 4),  or  4,  and  this  is  a  maximum  because  the  coefficient  of  x^ 
is  negative  ;  the  corresponding  value  of  x  is  2  h-  —  2,  or  —  1. 

647.  Let  a  be  a  value  of  x  that  makes  f{x)  a  maximum,  and  let 
h  be  a  small  positive  number,  as  small  as  we  please. 

Then,  however  small  h  is,  if /(.^•)  is  continuous,  by  definition 

/(a)  >f{a  ±  h).  (1) 

Therefore,  §  400,  if  k  is  any  constant, 

/(a)  +  k  >f(a  ±  h)  +  k;  that  is, 

Principle  2,  —  If  f(a)  is  a  maximum  of  f(x),  f(a)  +  k  is  a 
maximum  of  f(x)  -\-  k  ;  similarly,  if  f(a)  is  a  minimum  of  f{x), 
f(a)  +k  is  a  minimum  off(x)  -\-  k. 

Thus,  in  Fig.  2,  0  is  a  minimum  of  x^  —  8  x  +  16,  4  is  a  minimum  of 
x2  —  8  X  +  16  +  4,  or  x2  —  8  X  +  20,  and  —  4  is  a  minimum  of  x^— 8  x  +  16— 4, 
or  x2  -  8  X  +  12. 

Again,  if  f(a)  is  a  maximum  of  f(x),  by  (1)  and  the  principles 
of  inequalities, 
§402,  _/(a)<_Xa±/0; 

also,  §  403,  •      m/(a)  >  mf(a  ±  h) ; 

also,  §  403,  provided  /(a)  is  not  equal  to  zero,  dividing  (1)  by 

>  - — ,  or  < Hence, 

f{a±h)     f{a)         f(a)     f(a±h) 

Principles.  —  3.    If  f(a)  is  a  maximum  of  f(x),    —  f{a)  is  a    ' 
minimum  of  —  f{x). 

4.    Iffia)  is  a  maximum  of  fix),  mf(a)  is  a  maximum  ofmf(x). 


658  FUNCTIONS   OF  A    SINGLE    VARIABLE 

5.    If  f(a)  is  a  maximum  of  fix), is  a  minimum  of • 

JJK)  •''"'^'Aa)  ''fix) 

Thus,  4  is  a  maximum  of  3  —  2  x  —  x^  and  —  4  is  a  minimum  of  x^  +  2  x  —  3, 
as  may  be  discovered  by  applying  Prin.  1  or  by  plotting  their  graphs ;  also, 

8  is  a  maximum  of  2(3  —  2  x  —  x^)  and  \  is  a^^inimum  of 

The  same  principles  apply  to  minima  of  functions. 

648.  Let  f{x)  —  uv,  in  which  u  and  v  are  variables  whose  sum 
is  constant,  say  ?<  +  v  =  2  A". 

Since  (w  —  vf  is  positive  except  when  u  =  v,  from  the  identity 
4  MV  =  (m  -}-  ■v)^  —  {u  —  vY,  or  4  Mv  =  4  Tc^  —  (u  —  vy, 
4  uv  ^  4  /c^ ;  .-.  uv  ^  l<^.     Hence, 

Principle  6.  —  If  f(x)  is  the  product  of  tivo  variable  factors 
whose  sum  is  constant,  the  sqxiare  of  half  this  sum  is  a  maximum 

offi^y 

Since  3  —  2  x  —  x^  =  (3  +  x)(l  —  x),  the  sum  of  the  variable  factors  is  4,  a 
constant ;  then,  (|)2,  or  4,  is  a  maximum  of  the  function.     (See  Fig.  2.) 

Again,  to  find  the  critical  value  of  x^  —  8  x  +  12, 
put  x2  -  8  X  +  12  =  (x  -  6)(x  -  2)  =/(x). 

Then,  (x  -  6)(- x  +  2)  =-/(x). 

Therefore,  Prin.  6,        [|(x  —  6  —  x  +  2)]2,  or  4,  is  a  maximum  of  — /(x). 

Hence,  Prin.  3,  —  4  is  a  minimum  of /(x). 

In  geometrical  language  Prin.  6  is  stated  as  follows :  Of  all  rectangles  icith 
equal  perimeters,  the  square  is  the  greatest. 

649.  Let  f{x)  =  u  -\-  v,  in  which  u  and  v  are  variables  whose 
product  is  constant,  say  uv  =  Tc^. 

Since  (u  +  vy  =  4:  uv  +  (tt  —  vy,  and  since  (u  —  vy  is  positive 
except  when  u  =  v,  in  which  case  (u  +  v)^  =  4  uv, 

(u  +  v)2  ^  4  fc2 ;  that  is,  [f{x)y  ^  4 1^.  (1) 

Since  both  members  of  (1)  are  positive,  f(x)  is  numerically 
greater  than  ±  2  A;  or  else  equal  to  ±  2  k.  When  f(x)  is  positive, 
f(x)  ^  2  A; ;  and  when  f{x)  is  negative,  f{x)  -^  ~2k.     Hence, 

Principle  7.  —  If  f{x)  is  the  sum  oftivo  variables  whose  product 
is  constant,  twice  the  positive  square  root  of  this  product  is  a  mini- 
mum of  fix)  and  twice  the  negative  square  root  is  a  maximum. 

If  /(x)  =  =  X  +  -,  since  x  •  -  =  9,  a  constant,  6  is  a  minimum  and 

XXX 

—  6  a  maximum  of  the  function. 


FUNCTIONS   OF  A    SINGLE    VARIABLE 


559 


6. 


8. 


1  + 

X 

x  +  ^+2. 

X 


xj\        x) 


Examples 
Find  the  critical  values  of  the  following  functions : 

1.  rc^—  o  X  +  6. 

2.  x^+x  —  30. 

3.  3  ar- 4  a;  — 15.  7.    a; + 

4.  —a?  —  4. 

5.  (a;-3)(7-a;). 

change  of  sign  of 


650.    Critical  values  of  continuous  functions 
th^  first  derived  function. 

Since  the  first  derived  function  of  f{x)  is  the  limit  of  the  ratio 
of  the  increment  of  f(x)  to  the  increment  of  «  as  7i  =  0,  ^  being  a 
positive  increment,  it  follows  that  the  function  is  increasing  when 
f'{x)  is  positive  and  decreasing  when  fix)  is  negative,  and  vice 
versa. 

Hence,  if  the  derivative  off(x)  changes  sign  as  x  increases  through 
a,  f{a)  is  a  critical  value  of  the  function. 

The  value  of  a  is  found  by  solving  the  equation  /'(x)  =  0.     For  example, 
the  value  of  x  that  renders  a;^  _  g  x  +  12  a  minimum  is 
found    by    solving    2  a;  —  8  =  0,    giving  x  =  4,   whence, 
minimum  =/(4)  =  16  -  32  +  12  =  -  4.     (See  Fig.  2.) 

It  does  not  always  follow,  however,  that  the  values 
of  X  that  satisfy  the  equation  f'{x)  =  0  correspond  to 
critical  values  of  /(x).  For  the  derivative  may  become 
equal  to  zero  without  changing  sign  as  in  the  accom- 
panying graph  of  x^  —  1. 

661.   By  (5),  §  640,  when  x  =  a, 

fCa+  h)  -f(a)  =  hf'ia)  +|/"(a)  +||/"'(a)  +  - 

and  f(a  -  h)  -f(a)  =  -  hf'(a)  +^V"(a)  -||/"'(«)  + 


Suppose  that  y(a)  is  a  critical  value  oif(x).     Then,  §  650, 
f(a-\-h)-f(a) 


^|/"(«) +!>"(«)  + 


and         f(a  -  h)  -f{a)  =|/"(«)  -  ^/'"(a)  + 


560  FUNCTIONS   OF  A    SINGLE    VARIABLE 

If  h  is  taken  sufficiently  small,  the  terms  in  W,  h*,  •-■  may  be 
neglected.  Hence,  iij{a)  is  a  critical  value  oif(x),f"(a)  has  the 
same  sign  as  /(a  +  h)  —  J{a)  or  f(a  —  h)  —f(a).  But  by  the  test 
of  a  critical  value,  §  645,  f{a  +  h)  —f{a)  and  f{a  —  h)  —f(a) 
are  both  negative  when  f(ci)  is  a  maximum  and  both  positive  when 
f(a)  is  a  minimum. 

Hence,  if  the  first  derived  function  of  f(x)  changes  sign  when 
X  =  ci,  /(«)  is  a  maximum  or  a  minimum  of  f(x)  according  as  the 
second  derived  function  is  negative  or  positive  for  x  =  a. 

652.  Discontinuous  Functions. 

653.  When  the  increment  of  a  function  of  a  variable  corie- 
sponding  to  an  infinitesimal  increment  of  the  variable  is  not 
infinitesimal,  the  function  is  called  a  Discontinuous  Function. 

Let  h  be  any  increment  of  x  in  fix). 

Then,  f(x)  is  discontinuous  for  x  =  a,  if  the  difference  between 
f{a  +  h)  and  f{a  —  h)  cannot  be  made  as  small  as  we  please  by 
taking  h  sufficiently  small. 

The  function  x-\--  is   a  discontinuous 

X 

function  as  may  be  seen  from  its  graph. 
For  as  x  changes  from  a  very  small  nega- 
l  tive  number  to  a  very  small  positive  num- 
ber, the  function  changes  from  a  very  large 
negative  number  to  a  very  large  positive 
Y'  number;  and  the  smaller  the  increment  of 

ic  as  a;  passes  through  zero,  the  greater 
is  the  increment  of  the  function.  Hence,  the  function  is  discon- 
tinuous for  a;  =  0. 

For  all  other  values  of  x,  however,  the  function  is  continuous. 
The  graph  has  two  parts  called  branches,  one  in  the  third  quad- 
rant corresponding  to  negative  values  of  x,  the  other  in  the  first 
quadrant  corresponding  to  positive  values  of  x. 

654.  Let  h  and  k  be  corresponding  increments  of  x  and/(a;). 

If  the  ratio  -  is  always  finite  as  A  =  0,  fc  may  be  made  as  small 

as  we  please  by  taking  h  sufficiently  small ;  that  is,  the  function 
is  continuous  if  its  derivative  is  finite  for  all  values  of  x. 


X^ 


FUNCTIONS   OF  A    SINGLE    VARIABLE  561 

If  the  limit  of  this  ratio  can  become  infinite  for  some  value  of  x, 
an  infinitesimal  change  in  x  may  produce  a  finite  or  infinite 
change  in  f(x) ;  that  is,  the  function  is  discontinuous  for  x  =  a  if 
the  first  derivative  increases  or  decreases  without  limit  as  »  =  a. 

Thus,  X  +  i  is  discontinuous  for  x  =  0  ;  for  the  first  derivative  of  x  -f  x-'^ 

X 

is  1  —  a;-^,  or  1  —  -- ,  which  increases  or  decreases  without  limit  as  »  =  0. 

Examples 

1.    Discuss  the  function  a^  +  9  a^  +  23  x  + 15.     (See  Fig.  2.) 
Discussion.  — Since  the  function  is  rational  and  integral  with  respect  tox, 
it  is  continuous.     Tabulating  the  values  of  x  and  f{x)^ 


X    = 

—  00 

-6 

-5 

-4 

-3 

-2 

-1 

0 

1 

2 

+  00 

/(a^)= 

—  00 

-15 

0 

3 

0 

-3 

0 

15 

48 

105 

+  00 

This  shows  that  the  curve  begins  in  the  third  quadrant,  crosses  the  axis 
of  X  three  times,  giving  the  three  real  roots  —  5,  —  3,  and  —  1,  crosses  the 
axis  of  y  at  the  point  (0,  15),  and  extends  indefinitely  upward  in  the  first 
quadrant ;  also  that  the  function  is  an  increasing  function  up  to  some  value 
in  the  interval  between  x  =  —  5  and  x  =  —  3,  then  a  decreasing  function  to 
some  value  in  the  interval  between  x  =  —  3  and  x  =  —  1,  and  thereafter  an 
increasing  function.     To  find  the  turning  values,  put  /'(x)=  0. 

Then,  /'(x)=  3x2+ 18x  +  23  =  0. 

Solving,  X  =  —  3  ±  I V3  =  —  4.155  or  —  1.845,  approximately. 
Substituting  these  values  in  the  second  derived  function  6  x  +  18,  —  4.155 
makes  /"(x)  negative  and  —  1.845  makes /"(x)  positive. 

Hence,  §  051,  /(—  4.155)  is  a  maximum  and  /(—  1.845)  is  a  minimum. 
The  graph  may  now  be  plotted  completely. 

Discuss  and  plot  the  following  functions : 

2.  a?-6x^  +  llx-Q.  5.    ar'-7a;  +  6. 

3.  «2+l.  6.    ic*  — 5a^  +  4. 

4.  a^-10a;  +  25.  7.    x  +  4x-\ 

8.  Show  that  first  degree  functions  of  x  have  no  critical  values 
and  that  their  graphs  are  straight  lines. 

9.  Plot  the  graphs  of  a^  +  a^  —  5x-\-3  and  its  derived  func- 
tions. What  root  has  f(x)  =  0  when  f{x)  =  0  has  two  equal 
roots? 

ADV.  AlO.  —  36 


THEORY   OF   EQUATIONS 


655.  Changing  an  equation  to  another  whose  roots  are  the 
same  as  the  roots  of  the  given  equation  is  called  Reducing  the 
equation. 

656.  Changing  an  equation  to  another  whose  roots  have  known 
relations  to  the  roots  of  the  given  equation  is  called  Transforming 
the  equation. 

657.  Every  simple  equation  may  be  reduced  to  the  form 
ax-\-b  =  0',  every  quadratic  equation  to  the  form  aa^+6a;+c=0; 
every  equation  of  the  third  degree,  or  cubic  equation,  to  the  form 
aa^  -\-bx^  +  cx  +  d  =  0;  every  equation  of  the  fourth  degree,  or 
biquadratic  equation,  to  the  form  ax*  +  bxP  -{-cx^  +  dx  +  e  =  0;  etc. 

Hence,  these  equations  are  called  the  general  equations  of  the 
1st,  2d,  3d,  4th,  •••  degrees  in  x. 

The  general  equation  of  the  nth  degree  in  x  has  the  form 

a.T"  +  bx"-^  +  cx^--  H [-kx-\-l  =  0. 

658.  Dividing  both  members  of  the  general  equation  of  the 
nth  degree  in  x  by  the  coefficient  of  a;",  the  resulting  equation 
takes  the  form 

X"  -{-px"-^  +  qx''-^  -\ \-sx  +  t  =  0, 

in  which  the  coefficient  of  the  highest  power  of  x  is  1. 

This  equation  is  called  the  reduced  equation  of  the  nth  degree 
in  X.  The  first  member  is  a  rational  integral  function  of  x 
(§  638),  and  in  this  chapter  the  equation  will  often  be  written  in 
the  brief  form  J{x)  =  0. 

The  absolute  term  t  may  be  regarded  as  the  coefficient  of  x°. 

The  coefficients  p,  q,  •••,  s,  t  may  be  positive  or  negative  or 
equal  to  zero,  integral  or  fractional,  rational  or  irrational,  real 

562 


THEORY  OF  EQUATIONS  563 

or  imaginary.  But  unless  the  contrary  is  stated,  it  will  be  un- 
derstood that  they  are  rational  and  real. 

If  no  coefficient  is  equal  to  zero,  the  equation  is  complete;  if 
any  coefficient  is  equal  to  zero,  the  equation  is  incomplete. 

659.  Divisibility  of  ^(jr)  =  0. 

Let  the  first  member  of  /(a;)  =  0  be  divided  hy  x  —  a  until  the 
remainder  no  longer  involves  x. 

Denote  the  quotient  by  Q  and  the  remainder  by  R. 

Then,  f{x)  =  (x-a)Q  +  B.  (1) 

Since  M  does  not  involve  x,  R  has  the  same  value  whatever 
value  X  has.     To  find  R,  then,  let  x  =  a. 

When  a  is  substituted  for  x,  f{x)  becomes  /(a)  by  the  defini- 
tion of  /(a)  ;  and  the  second  member  of  (1)  becomes  0  •  Q  +  jR, 

ox  R. 

.-.  f(a)  =  R;  that  is,  i?=/(a).     Hence, 

Principle  1.  —  If  f(x)  is  divided  hy  x—a,  the  remainder  is 
f(a).     Consequently, 

Corollary  1.  —  If  f(a)  =  0,  that  is,  if  a  is  a  root  of  the  equa- 
tion f(x)  =  0,  f(x)  is  divisible  by  x  —  a  (Factor  Theorem),  and  con- 
versely, if  f(x)  is  divisible  by  x  —  a,  then  f{a)  =  0  and  a  is  a  root 
of  the  equation  f(x)  =  0. 

Principle  1  is  sometimes  called  the  Remainder  Theorem.  The 
Factor  Theorem,  §  136,  is  a  special  case  of  the  Remainder 
Theorem. 

If  the  coefficients  of  x"-  +^xb""^  +  ga;""- -| \-sx-\-tz=0  are  all 

integers,  the  first  member  cannot  be  divisible  by  x  —  a  unless  a 
is  a  factor  of  t.     Hence, 

Corollary  2.  —  If  f(x)  has  integral  coefficients,  every  integral 
root  of  f(x)  =  0  must  be  a  factor  of  the  absolute  term. 

660.  Number  of  roots. 

Since  by  Prin.  1,  Cor.  1,  to  every  root  of  the  equation  f(x)  =  0 
there  corresponds  a  factor  of  f(x)  of  the  form  x  —  a,  and,  con- 
versely, to  every  factor  of  this  form  there  corresponds  a  root  of 
the  equation  f(x)  —  0,  it  may  be  inferred  that : 

Principle  2.  —  Every  equation  of  the  nth  degree  in  x  has  n  and 
only  n  roots. 


564  THEORY  OF  EQUATIONS 

This  principle  cannot  be  proved  until  it  has  been  proved  that  every  equa- 
tion has  at  least  one  root  —  a  fact  too  difficult  of  proof  for  this  work.  But 
assuming  that  every  equation  has  at  least  one  root,  Prin.  2  may  be  estab- 
lished conditionally  as  follows : 

If  fix)  =  x»  +  px"-!  +  gx»  2  ^  ...  4.  sx  4- «  =  0  has  a  root,  let  it  be  a. 

Dividing  both  members  by  a;  —  a,  an  equation  of  the  (n  —  l)th  degree  is 
obtained.  If  this  has  a  root,  let  it  be  h.  Dividing  both  members  by  a;  —  6, 
an  equation  of  the  (n  —  2)th  degree  is  obtained.  By  continuing  this  process 
/(x)  is  resolved  into  n  factors  of  the  form  x  —  a.    Denoting  these  by  x  —  a, 

f{x)={x  -  a)(x-  b)  •'•  (x  -  k). 

Since  the  substitution  of  any  one  of  the  n  numbers  a,  6,  •••,  k  will  reduce 
/(x)  to  ?ero,  the  equation  f(x)  =  0  has  these  n  numbers  for  roots.  And 
since  the  substitution  of  any  other  number  than  one  of  these  will  not  reduce 
/(x)  to  zero,  /(x)  =  0  has  no  other  roots. 

Note.  —  In  counting  the  roots,  equal  roots  are  regarded  as  different  roots. 

661.  If  a  is  a  root  of  the  equation  f(x)  =  0,  then,  by  Prin.  1, 
Cor.  1,  fix)  is  divisible  by  a;  —  a.  Removing  the  root  a  by  divid- 
ing both  members  of  f{x)  =  0  by  a?  —  a  is  called  depressing  the 
equation  fix)  =  0  to  an  equation  of  the  next  lower  degree,  and 
this  equation  is  called  the  depressed  equation. 

662.  Horner's  method  of  synthetic  division. 

Horner's  method  of  synthetic  division  is  an  abridgment  of  the 
method  of  division  by  detached  coefficients. 

In  dividing  5.'c'+8x*+12a^+24a^-5x+7  by.ar'+2a^+3ic+5, 

the  process  of  division  by  detached  coefficients  is  modified  as 

follows : 

Dividend 
Divisor 

with  last  three 
signs   changed 

Quotient  5  —    2  +   l|        3+    2  +  2     Remainder 

Explanation.  — By  using  the  first  term  of  the  divisor  as  a  trial  divisor 
and  changing  the  sign  of  each  term  of  the  divisor  ailer  the  first,  it  is  possible 
to  substitute  for  the  process  of  subtracting  the  partial  products  that  of  adding 
the  products  obtained  by  multiplying  the  terms  of  the  divisor  after  the  first, 
icith  their  signs  changed,  by  the  successive  terms  of  the  quotient. 

Thus,  dividing  the  coefficient  of  the  first  term  of  the  dividend  by  the  coeffi- 
cient of  the  first  term  of  the  divisor,  5  is  obtained  for  the  coefficient  of  the 


1 

5+   8  +  12  +  24-   5  +  7 

2 

-10-15-25 

3 

+   4+    6  +  10 

5 

-    2-   3-5 

THEORY  OF  EQUATIONS  '  565 

first  term  of  the  quotient.  Multiplying  the  coefficients  of  the  other  terms  of 
the  divisor  with  their  signs  changed  by  5  gives  the  second  horizontal  line  of 
the  process,  —  10  —  15  —  25,  which  is  added  to  the  dividend.  At  present, 
however,  only  the  term  —  10  is  added. 

Adding  +  8  and  —  10,  the  terms  in  the  second  column,  and  dividing  the 
sum  by  1,  the  coefficient  of  the  first  term  of  the  divisor,  —  2  is  obtained  for 
the  coefficient  of  the  next  term  of  the  quotient.  Multiplying  the  coefficients 
of  the  last  three  terms  of  the  modified  divisor  by  this  term  of  the  quotient 
gives  the  third  horizontal  line  of  the  process. 

Adding  +  12,  —  15,  and  +  4,  the  terms  in  the  third  column,  and  dividing 
the  sum  by  the  coefficient  of  the  first  term  of  the  divisor,  +  1  is  obtained  for 
the  coefficient  of  the  next  term  of  the  quotient.  Multiplying  the  coefficients 
of  the  last  three  terms  of  the  modified  divisor  by  this  term  of  the  quotient 
gives  the  fourth  horizontal  line  of  the  process. 

Adding  +  24,  —  25,  +  6,  and  —  2,  the  terms  in  the  fourth  column,  the 
sum  is  3,  which  stands  for  3  a;2.  Since  this  term  is  of  lower  degree  than  the 
first  term  of  the  divisor,  5  x^  —  2  a;  +  1  is  the  entire  quotient,  and  3  or?  is 
the  first  term  of  the  remainder.  Adding  the  last  two  columns,  the  entire 
remainder  is  3  x'-'  +  2  x  +  2. 

663.  When  the  divisor  is  a  binomial  of  the  form  x-^a,  the 
first  term  is  not  written,  and  the  second  term  with  its  sign  changed 
is  written  at  the  right  of  the  dividend.  Also,  since  each  partial 
product  consists  of  but  one  term,  all  the  partial  products  are 
written  in  the  same  line  under  the  dividend.  For  example,  the 
process  of  dividing  a^  +  4a^  +  7a;  +  6  by  x  —  2  is  as  follows : 

-Dividend  1  +  4+   7+   6L2 

Partial  products  2  +  12  +  38 

Quotient  1  +  6  +  19  I  44    Remainder 

That  is,  the  quotient  is  o:^  +  6  ic  +  19  and  the  remainder  44. 

It  is  seen  that  the  process  of  dividing  f(x)  by  x  —  2  is  identical 
with  the  process  of  substituting  2  for  x  by  detached  coefficients 
(§  628),  the  remainder  being  the  value  of  the  function  when  x=2. 

This  illustrates  the  meaning  of  Prin.  1. 

Examples 
Divide  by  synthetic  division : 

1.  ic*  +  a^  -  3  a^  -  17  a;  -  30  by  «  -  3 ;  by  a^  +  2. 

2.  af  +  4a^  —  Bar'  —  a;  +  2bya;  —  1;  bya;  +  l. 


566  THEORY   OF  EQUATIONS 

3.  2  X*  —  7  ic^  —  16  ar'^  —  ic-  +-  32  X  —  10  by  ic  —  5 ;  by  a;  —  1. 

4.  2a;5_5^4_^4,p3_22a;  +  21  by  ar-2a;  +  3. 

664.    To  find  the  commensurable  roots  of  f{x)  =  0  by  trial. 

Examples 
1.    Solve  f{x)  =  a:^  -  8  x*  +- 15  a.-3  +  20  0^  -  76  a;  +  48  =  0. 

SOLUTIOK 


1 

-  8  +  15  +  20  -  76  +  48  Li 

1  _    7  +    8  +  28-48 

1- 

-  7  +    8  +  28-48 
2  _  10  -    4  +  48 

L2 

1 

-5-    2  +  24 
-2  +  14-24 

1-2 

1-7  +  12 
That  is,  a;2  -  7  a:  +  12  =  0,  or  (x  -  3)(x  -  4)  =  0. 

.-.  X  =  1,  2,  -  2,  3,  4. 

Explanation. — By  Prin.  1,  Cor.  2,  every  integral  root  of  the  equation 

must  be  a  factor  of  48.  Substituting  1  for  x  the  result  is  zero,  and  therefore 
1  is  a  root  of  the  equation ;  also  since  the  process  of  substituting  1  for  x  is 
identical  with  that  used  to  divide  /(x)  by  x  —  1,  the  remainder  is  zero,  and 
the  depressed  equation  is  x*  —  7  x^  +  8  x'^  +  28  x  —  48.  This  equation  must 
have  the  other  roots  of  the  given  equation. 

Similarly,  removing  the  root  2  from  the  depressed  equation,  the  next 
depressed  equation  is  x^  —  5  x^  —  2  x  +  24  =  0.  Removing  the  root  —  2  from 
this  equation,  the  quadratic  x^  —  7  x  +  12  =  0  is  obtained,  whose  roots  are  3 
and  4. 

Hence,  the  roots  of  /(x)  =  0  are  1,2,  —  2,  3,  and  4. 

Solve  by  trial : 

2.  a;3-da^  +  23a;-15  =  0.  4.    :^-7x  +  Q  =  Q>. 

3.  a;3_i0x^  +  29a;-20  =  0.  5.   a;*-9x2  +  4x-f  12  =  0. 

6.  a;3-10x2_^33^_3e^0. 

7.  x-*-9a^4-21a:'+-a;-30  =  0. 

8.  5x*~2x'-^ox'-lQx  +  12  =  Q. 

9.  2a^-x*-12:»?-\-lx'  +  lQx-12==^{i. 


THEORY  OF  EQUATIONS  567 

665.   Newton's  method  of  divisors. 

The  following  method  of  limiting  the  number  of  factors  of  the 
absolute  term  to  be  tried  for  roots  is  rarely  expedient  but  is  useful 
as  a  check. 

Let  a  be  an  integral  root  of  the  equation 

y(x)  =  a;"  +  pyf-^  +  qx""'^  +  •••  +  rx^  +  sx  +  t  =  0, 
and  suppose  that  all  the  coefficients  are  integers. 
Then,       a»  +  pa"-^  +  Q-a""^  +  •••  +  ra-  +  sa-\-t  =  0. 
Transposing  and  dividing  by  a, 


-  =  —  s  —  ra  —  ••••  —  qa"-"  —pa""'^  —  a" 
a 


,»-3  /r.«»-2 


Hence,  -  is  an  integer. 
a 

Denoting  the  quotient  by  Qj,  and  transposing  —  s, 

Qi  +  s  =  —  ra—  •••  —  qa"~^  —  pa"  ~^  —  a"~\ 
Dividing  again  by  a, 

V^ "'  ^  —  —  r  —  •••  —  qa'*-*  —pa"~^  —  a"~^,  an  integer. 
a 

Denoting  this  quotient  by  Q^  and  continuing  the  process,  the 
result  of  the  nth  division  is 

Qn-l  +  P  =  _  1, 
a 

Let  d  be  an  exact  divisor  of  the  absolute  term  of  f(x)  =  0. 
Then,  by  the  above  discussion  we  have  the  following : 

Rule.  —  Divide  tJie  absolute  term  hjy  d^and  add  the  quotient  to  the 
coefficient  of  x;  divide  the  sum  by  d,  and  if  the  quotient  is  an  integer, 
add  it  to  the  coefficient  of  x? ;  continue  this  process  as  long  as  the 
quotients  are  integers  or  to  n  divisions. 

If  each  quotient  is  integral  and  the  nth  quotient  is  —  1,  then  d  is  a 
root  of  the  equation ;  'oih&nmse  d  is  not  a  root. 

Thus,  —  2  is  a  root  ot  x^  —  3  x^  —  16  x  —  12  =  0,  but  3  and  2  are  not  roots, 
as  shown  by  the  following  processes : 


1_  3-16-12  L 
-1+5+    9 

-2 

1  -  3  -  16  - 
-    4 

-12  L3 

1  -    3  -  16  -  12  [2 
-7-11-    6 

0  +  2-10 

-20 

-  14  -  22 

568 


THEORY  OF  EQUATIONS 


666.  Relation  between  the  roots  and  coefficients  of  an  equation. 

Since  the  reduced  equation  of  the  nth  degree  in  x  is  the  product 
of  n  equations  of  the  first  degree  of  the  form  a;  —  a  =  0,  an  equa- 
tion may  be  formed  from  its  roots  by  subtracting  each  from  x  and 
equating  to  zero  the  product  of  the  n  binomials  thus  formed. 

Thus,  suppose  x  =  a,  b,  c,  d. 

Then,      x  —  a  =  0,  x  —  b  =  0,  x  —  c  =  0,  x  —  d  =  0. 

.-.  (x  —  a) (x  —  b)(x  —  c) (x  —  d)=0, 


or 


—  a 

a?  -\-  ah 

y?  —  abc 

-  b 

+  ac 

—  abd 

—  c 

-\-ad 

—  acd 

-d 

-f  be 

-bed 

+  bd 

+  cd 

X  +  abed  =  0. 


Comparing  the  coefficients  of  the  above  equation  with  its  roots 
a,  b,  c,  d,  with  their  signs  changed,  the  following  rule  for  finding 
the  coefficients  when  the  roots  are  given  is  obtained : 

Rule.  —  Tlie  coeffieient  of  the  first  term  of  fix)  =0  is  1 ;  the 
coefficient  of  the  second  term  is  the  sum  of  the  roots  with  their  signs 
changed;  the  coefficient  of  the  third  term  is  the  sum  of  the  products 
of  the  roots  with  their  signs  changed,  taken  two  at  a  time;  and,  in 
general,  the  coefficient  of  the  (r  +  1)^^  term  is  the  sum  of  the  products 
of  the  roots  with  their  signs  changed,  taken  r  at  a  time. 

The  absolute  term  is  the  product  of  all  the  roots  with  their  signs 
changed. 

1.  If  the  second  term  is  wanting,  the  sum  of  the  roots  is  0. 

2.  If  the  absolute  term  is  wanting,  one  root  at  least  is  0. 


Examples 
Form  the  equations  whose  roots  are : 

1.  1,  2,  3.  3.    - 1,  -  3,       4. 

2.  2,  3,  5.  4.        1,       1,  -  f 


5.  2,  -|±V5. 

6.  1,    -^±^y/^3. 


667.  It  has  been  shown  that  the  relations  subsisting  between 
the  roots  and  coefficients  of  an  equation  may  be  expressed  by  n 
■conditional  equations,  sufficient  to  determine  the  coefficients  when 


THEORY  OF  EQUATIONS  569 

the  roots  are  known.  It  might  be  supposed,  therefore,  that  when 
the  coefficients  are  known  the  roots  may  be  found  by  aid  of  their 
relations  to  the  coefficients.  It  is  not  advantageous,  however,  to 
endeavor  to  find  the  roots  in  this  way,  except  when  some  particu- 
lar relation  between  the  roots  is  given  in  addition. 

Thus,  if  the  roots  of  x*  +  px-  +  qx  +  r  =  0  are  to  be  found  by  means  of 
their  relations  to  the  coefficients  p,  q,  and  r,  denoting  the  roots  by  a,  6,  and  c, 

a  +  b  +  c  =  -p, 
ab  +  ac+  be  =  q, 
abc  =—  r. 

Eliminating  any  two  of  the  three  unknown  numbers,  as  b  and  c,  it  is  found 
that  the  solution  of  the  system  depends  upon  the  solution  of  an  equation  of 
the  form 

a^  +  pcfi  +  5'a  +  r  =  0, 

which  is  as  difficult  to  solve  as  the  given  equation,  being  of  the  same  form. 

The  following  examples  illustrate  the  method  of  solving  an 
equation  by  means  of  the  general  relations  between  its  roots  and 
coefficients  when  the  roots  also  have  some  given  relation. 

Examples 

1.  Solve  the  equation  48  a^ —  74  a?^  +  37  a;  —  6  =  0,  whose  roots 
are  in  geometrical  progression. 


Solution.  —  Represent 

the  roots  by  - ,  a,  and  ar. 
r 

Then,  §  666, 

a                          74 

---  a-  ar  =  -  —  , 

r                          48 

(1) 

d 

r                        48 

(2) 

Dividing  (2)  by  (1), 

—  a  =  —  i,  or  a 

=  h 

(3) 

Substituting  in  (1), 

r  =  1  or  f . 

(4) 

Testing  these  values  in  the  third  equation  of  condition  by  which  the  abso- 
lute term  is  found  from  the  roots, 

(-»)(-„K-a0.or-a.=-A  =  ^(i)', 

by  (3)  they  satisfy  this  relation. 

Hence,  the  roots  are         f ,  ^,  f ,  when  r  =  f ; 
or  f ,  J,  §,  when  r  =  f . 


570  THEORY  OF  EQUATIONS 

2.  Solve  9a^— 7a;  +  2  =  0,  if  one  root  is  double  another. 

3.  Solve  9  a^  —  27  a;^  +  23  a;  —  5  =  0,  whose  roots  are  in  A. P. 

4.  Solve  the  equation  a^  —  12  a;  +  16  =  0,  which  has  two  equal 
roots. 

5.  Solve  2  a;*  —  7  .T^  —  6  a^  +  44  a;  —  40  =  0,  which  has  three 
equal  roots. 

6.  Solve  72a;*+90a^-5a^- 40a;- 12  =  0,  two  of  whose 
roots  are  numerically  equal  but  opposite  in  sign. 

7.  Solve   6a^  — lla^  +  6a;  — 1  =  0,  whose  roots  are  in  H.P, 

8.  Find  the  relation  subsisting  between  the  coefficients  of  the 
equation  aj*  +  pa?  -{-  qx-\-r  =  Q,  when  two  roots  are  numerically 
equal  but  opposite  in  sign,  and  by  the  aid  of  this  relation  solve 
the  equation  a:^  —  5  a;^  —  4  a;  +  20  =  0. 

668.  Fractional  roots. 

If  possible,  suppose  that  -,  a  rational  fraction  in  its  lowest 

h 

terms,  is  a  root  of  the  equatioirf* 

f{x)  =  a;"  +  29a;»-i  +  q^-^  -\ 1-  sa;  +  ^  =  0, 

whose  coefficients  1,  p,  q,  •••,  s,  t  are  integers. 

Substituting  this  root  for  x  and  multiplying  both  members 

by  6"-\ 

—  4-i)a"~^  +  qa^'-^b  -\ h  sa6»-2  +  tb"-'^  =  0 ; 

b 

whence,     ^  =  —  (pa"-^  +  qa^'-^b  H +  sab""-^  +  tb"'^), 

b 

which  is  impossible,  because  the  first  member  is  a  fraction  in  its 
lowest  terms  and  the  second  member  is  an  integer.     Hence, 

Principle  3.  —  If  the  coefficient  of  the  highest  power  of  x  in 
f(x)  =  0  IS  1  and  all  the  other  coefficients  are  integers,  no  rational 
fraction  can  be  a  root  of  the  equation. 

669.  Imaginary  roots  and  quadratic  factors  of  f(x). 

In  the  following  pages  every  complex  or  pure  imaginary  root,  that  is, 
every  root  involving  an  imaginary  number,  will  be  called  an  imaginary  root. 

Let  a  4-  6  V—  1  and  a  —bV—1,  in  which  a  and  b  are  real,  b 
is  positive,  and  a  is  positive  or  zero,  be  substituted  for  x  in  an 
equation  f(x)  =  0  having  positive  integral  exponents  and  real 
coefficients. 


THEORY  OF  EQUATIONS  571 

Theu,  j\a  +  6  V^l)  =  (a  +  6  V^n[)" +i:'(a  +  6  V^H^)""^  H 

+  s(a  +  6V^^)  +  ^,  (1) 

and  /(a  -  b  V^^)  =  {a-h  V^H!)"  +  p{a-b  V^^)""'  +  ••• 

+  s(a-6V=^)  +  f.  (2) 

Since  all  the  exponents  are  positive  integers,  (1)  and  (2)  may 
be  fully  expanded.  Each  will  then  consist  of  terms  of  three 
kinds,  those  which  do  not  involve  6,  those  which  involve 
even  powers  of  6V— 1,  and  those  which  involve  odd  powers 
of  6  V  —  1.  Terms  belonging  to  the  first  and  second  classes  will 
be  real,  and  to  the  third  class,  imaginary. 

Since  the  two  complex  numbers  are  alike  except  in  the  terms 
involving  h,  the  sum  of  the  terms  that  do  not  involve  h  is  the 
same  in  (2)  as  in  (1).     Represent  this  sum  by  S. 

Since  any  even  power  of  —  6  V—  1  is  equal  to  the  same  power 
of  6  V—  1,  the  sum  of  the  terms  involving  even  powers  of  6  V—  1 
is  the  same  in  (2)  as  in  (1).     Represent  this  sum  by  P. 

Since  any  odd  power  of  —  6  V  —  1  is  numerically  equal  to  the 
same  power  of  6  V—  1  but  opposite  in  sign,  if  the  sum  of  those 
terms  that  involve  odd  powers  of  6  V—  1  in  (1)  is  denoted  by 
Q  V—  1,  the  sum  of  the  corresponding  terms  in  (2)  is  —  Q  V—  1. 

Hence,  /(a  +  b  V^^)  =  S  +  P+Q  V^  (3) 

and  /(a  -  b  V^T)  =  S  +  P-Q  V^l.  (4) 

Now  suppose  that  one  of  the  conjugate  complex  or  imaginary 

numbers,  as  a  +  6V—  1,  is  a  root  of  the  equation  f{x)  =  0. 
Then,  S  +  P+Q  V^^  =  0.  (5) 

Since  the  real   number  (S  +  P)  can   cancel   no   part   of  the 

imaginary  number  Q  V—  1,  each  must  be  equal  to  zero. 

Therefore,  S  +  P-Q  V^^  =  0 ;  (6) 

that  is,  a  —  &  V—  1  also  is  a  root  of  the  equation.     Hence, 

Pbinciple  4.  —  If  the  equation  f(x)  =  0  has  real  coefficients, 
its  imaginai-y  roots,  if  any,  occur  in  conjugate  pairs. 

CoROLLARiKS.  —  1.  To  every  pair  of  imaginary  roots  there  corre- 
sponds a  quadratic  factor  of  f(x)  of  the  form  (x  —  aY  +  h^. 

2.  If  the  equation  f{x)  =■  0  is  of  an  odd  degree,  it  has  at  least 
one  real  root. 


572  THEORY  OF  EQUATIONS 

Examples 

1.  One  of  the  roots  of  the  equation  ic*— 2a^+10ar^— 8a;+24=0 
isl+V— 5.     Solve  the  equation. 

Solution.  —  Since  the  coefficients  are  all  real,  the  given  root  must  enter 
the  equation  with  its  conjugate  in  the  quadratic  factor 

(x  -  1  -  \/^5)  (x  -  1  +  V^),  or  x^  -  2 X  +  6. 

Removing  this  factor  by  synthetic  division,  the  depressed  equation  is 
x2  +  4  =  0,  whose  roots  are  2  V—  1  and  —  2  V—  1. 

Hence,  the  roots  of  the  given  equation  are  1  ±  V—  5,  ±  2^—  1. 

Solve  the  following,  having  given  the  roots  indicated : 

2.  2ar'-lla^.+  8a;  +  7  =  0;3+V2. 

3.  2a^-lla^  +  16a;  +  ll  =  0;3-V^2. 

4.  a;<-2a^-ar=  +  6a;-3  =  0;  -i(3  +  V^=^). 

5.  ic*  +  3ar'  +  5a^  +  4a;  +  2=0;   -1(1+^/"^). 

6.  2ar>-a;^  +  4a^-2a;2  +  2a;-l  =  0;  V^  V^^. 

TRANSFORMATION  OF  EQUATIONS 

670.  Frequently  by  transforming  an  equation  its  solution  may 
be  made  to  depend  upon  that  of  an  equation  that  is  easier  to  solve. 

For  example,  it  is  easier  to  substitute  positive  integral  values 
of  the  unknown  number  than  negative  or  fractional  values. 

Hence,  to  find  the  negative  roots  of  f{x)  =  0  by  trial,  first  trans- 
form to  an  equation  whose  roots  are  numerically  equal  to  the  roots 
of  fix)  =  0  but  opposite  in  sign,  and  then  search  for  the  positive 
roots  of  the  transformed  equation. 

671.  To  change  the  signs  of  the  roots. 

Given       f{x)  =  a;"  +  px"-^  +  qx''-^  -\ \-sx-\-t  =  0.  (1) 

Put  —  y  for  x.  Then,  the  terms  of  (1)  involving  odd  powers 
will  have  their  signs  changed  and 

yn  _  pyn-l  _^  gyn-2 sy  +  t  =  0  (2) 

or  -  y+py"-'  -  92/"-'  -\ sy  -\-t  =  0,  (3) 

according  as  n  is  even  or  odd.     Changing  all  signs  in  (3), 

yn  _  pyn-i  ^  gyn-2 _^gy_t=,0.  (4) 


THEORY  OF  EQUATIONS  573 

Since  —y  =  x,y  =  —  x  and  the  roots  of  (2)  and  (4)  are  those  of 
(1)  with  their  signs  changed.  (2)  is  obtained  by  changing  the 
signs  of  odd  powers  and  (4)  by  changing  the  signs  of  even  powers. 

Principle  5.  '—  An  equation  in  x  may  he  transformed  into  another 
having  the  same  roots  ivith  opposite  signs  by  changing  the  signs  of 
the  terms  involving  either  the  odd  or  the  even  powers. 

To  change  the  signs  of  the  roots  ofar*  +  x3  +  2x  +  3  =  0  and  of  x^  +  2 x^ 
+  4  a;2  +  X  +  5  =  0,  and  leave  the  highest  power  positive,  change  the  signs 
of  odd  powers  in  the  first  instance  and  of  even  powers  in  the  second.  The 
transformed  equations  are,  using  y  for  the  changed  roots, 

y4_y3_2y  +  3  =  0  and  y^  +  2  y^  -  ^y'^  +  y  -  b  =  0. 

The  absolute  terras,  3  and  5,  or  3  x"  and  5  x°,  are  regarded  as  even  powers. 

672.  To  multiply  the  roots  by  a  constant. 

Given        fix)  —  a;"  +  px""-^  +  gx"--  H 1-  sa;  +  <  =  0,  (1) 

and  m  a  constant  by  which  each  root  is  to  be  multiplied. 

Let  y  =  mx,  (2) 

whence,  x  =—•     Substituting  —  for  x  in  (1)  and  multiplying  both 
members  of  the  resulting  equation  by  m", 

2/"  4-  mpy""-^  +  m^qy''~^  +  •  •  •  -f  m'^'^sy  +  m^t  =  0,  (3) 

whose  roots  by  (2)  are  m  tiines  those  of  (1).     Hence, 

Principle  6.  —  An  equation  f(x)  =  0  may  be  transformed  into 
another  whose  roots  are  m  times  those  of  the  given  equation  by  multi- 
plying the  successive  coefficients  by  1,  m,  m^,  m%  •••,  m",  respectively. 

To  multiply  the  roots  of  x^  —  6  x^  +  11  x  —  6  =  0  by  2,  the  successive  coef- 
ficients are  multiplied  by  1,  2,  4,  8,  respectively,  and  y  or  some  other  letter  is 
used  for  the  unknown  number  in  the  transformed  equation.     Hence,  the  equa- 
tion whose  roots  are  twice  those  of  the  given  equation  is 
ys  _i2yi  +  Uy  -48  =  0. 

673.  The  principal  use  of  the  preceding  transformation  is  to 
clear  equations  of  fractional  coefficients  and  at  the  same  time  make 
the  coefficient  of  the  highest  power  of  x  unity. 

If  x2  _  |a;  -I-  3  =  0  is  cleared  of  fractional  coefficients  without  changing  the 
roots  f  and  2,  the  coefficient  of  x^  in  the  resulting  equation  will  not  be  1, 
But  multiplying  the  roots  by  2,  giving  y'^-ly  +  12  =  0,  the  equation  is  cleared 
of  fractional  coefficients  and  the  coefficient  of  the  highest  power  of  the 
unknown  number  is  still  1. 


574  THEORY  OF  EQUATIONS 

Examples 

Transform  the  following  equations  to  equations  whose  roots 
are  ten  times  the  roots  of  the  given  equations : 

1.  x'+2x-+x-4.=0.  3.    ar'-.9a-2+. 23a; -.015=0. 

2.  a^-Sar  -4:X+12=0.  4.    x*-.18a^+. 032 a; -.0015=0. 
5.    Transform  8  .r'*  —  12  cc^  +  4  a;-  —  10  .t  —  3  =  0  to  an  equation 

with  integral  coefficients,  the  leading  coefficient  being  1. 


Solution 

Dividing  by 

8, 

a;*- 

-fa:3  + 

ix'- 

■ix 

-i-- 

=  0. 

Multiplying 

the  roots 

;  by  m 

,  Prin.  6, 

2/4- 

2   ^ 

^fy- 

5m^ 

2 

y  - 

3to4 

8 

=  0, 

whose  coefficients  are  evidently  integral  when  m  =  2. 

Substituting  2  for  »n,      y*  -  3  yS  +  2  j/'^  -  20  ?/  -  6  =  0, 
an  equation  whose  roots  are  double  those  of  the  given  equation. 

Transform  the  following  equations  to  the  reduced  form/(a;)=0 
having  integral  coefficients  • 

6.  4a;*+3a^-2a;+9=0.  9.    7?-\x'+j\x-J^=0. 

7.  16 0^3-12 ar'+10a;-7=0.     10.    (a;-lf(3.T-l)2=50. 

8.  2a^— V-a^+-V- 2^-11= 0-    H-    5a;^+.la^+.125x+.2=0. 

12.  On  page  299  it  is  shown  that  the  three  cube  roots  of  1  are 
1,  ^  (—  1  ±  V— 3).  Prove  by  Prin.  6  that  the  three  cube  roots  of 
a?  are  a,  ia(— 1  ±  V— 3). 

Find  the  three  cube  roots  of  8,  27,  —  8,  and  —  1. 

674.   To  decrease  the  roots  by  a  given  number. 

Given        J{x)  =  .r"  +  p.T"-'  +  qx"--  +  •  ■  +  sx  +  t  =  0,  (1) 

and  let  it  be  required  to  find  an  equation  of  the  same  form  whose 
roots  are  h  less. 

First  Method.         Let  y  =:x  —  h,  whence  x  =  y  +  h. 

Substituting  (y  +  h)  for  a;  in  (1), 
(?/  4-  /O"  +  PQ^  +  hy-'  +  g(y  +  h)"-'  +  •  •  •  4-  s(y  +  J>)  +  f  =  0.     (2) 

Since  y  =  x  —  h,  (2)  is  the  required  transformed  equation  before 
reduction  to  the  same  form  as  (1). 


THEORY  OF  EQUATIONS 


575 


Expanding  the  powers  of  {y  -j-  h)  in  (2)  and  arranging  the  result 
in  descending  powers  of  y, 


y'^  +  nh  y«-'  +  ^ — 11 /i2 


+   P 


+   (n-  l)pli 

+  q 


2/»-2.j 1_      /i« 


+ 
+ 


+  sh 
+     t 


=  0, 


(3) 


which  is  the  reduced  form  of  the  transformed  equation. 

Second  Method.  Except  when  n  is  small  the  expansion  of  the 
powers  of  {y  +  h)  is  laborious.  To  discover  an  easier  method  of 
finding  the  coefficients  of  2/"~S  y"~^  •••in  the  transformed  equation 
(3),  represent  these  coefficients  by  P,  Q,  •••. 

Then,  ?/"  +  Py""^  +  Q?/"-^  -\ \-Sy+T=0  (4) 

is  an  equation  obtained  from  (1)  by  substituting  y  +  h  for  x,  or 
y  for  X  —  h.  Therefore,  if  a;  —  ^  is  substituted  for  y  in  (4),  the 
first  member  of  the  resulting  equation  must  be  identically  equal 
to  the  first  member  of  (1) ;  that  is, 

a;"  +  px"-^  4-  qx'"-^  -\ \- sx  +  t  =  {x  —  hy -\-  P(x  —  h)"-^ 

+  Q{x  -  hy-'  +...+S{x-  h)  +  T  (5) 

is  an  identical  equation. 

From  the  form  to  which  we  have  reduced  f(x)  in  the  second 
member  of  (5),  it  is  evident  that  T,  the  last  coefficient  of  the 
transformed  equation,  is  the  remainder  when  f{x)  is  divided  by 
X  —  h;  S,  the  preceding  coefficient,  is  the  remainder  when  the 
quotient  is  divided  hy  x  —  h;  etc. 

Hence,  to  transform  an  equation  of  the  nth  degree  in  x  into 
another  whose  roots  are  h  less, 

Rule.  —  Substitute  y  —  h  for  x  in  the  given  equation  and  reduce 
the  resulting  equoMon. 

Or,  divide  f{x)  hy  x  —  h  until  the  remainder  does  not  involve  x, 
divide  the  quotient  by  x  —  h  until  the  remainder  does  not  involve  x, 
and  so  continue  to  n  divisions.  Tlie  first  coefficient  of  the  given 
equation  together  with  the  successive  remainders  taken  in  the  reverse 
order  will  be  the  coefficients  of  the  transformed  equation. 


676  THEORY  OF  EQUATIONS 

Examples 
1.   Decrease  the  roots  of  2  ar*  -  19  a^  +  59  a;  —  60  =  0  by  2. 
First  Solution.  —  Let  x  -  2  =  y.     Then,  x  =  y  +  2.     Substituting, 

2(2/ +  2)3 -19(2/ +  2)2 +  59(2/ +  2) -60  =  0.  (1) 

Reducing,  2  j/S  -  7  r/^  +  7  2/  -  2  =  0.  (2) 

Skcond  Solution.  —  Dividing  2x3  —  19  x^  -).  59  a;  —  60  by  x  —  2  as  in  the 
following  process,  the  quotient  is  2  x^  -  15  x  +  29  and  the  remainder  is  -2. 

Dividing  2  x^- 15  x+29  by  x-2,  the  quotient  is  2  x- 11  and  the  remainder 
is  7. 

Dividing  2x  —  llbyx  —  2,  the  quotient  is  2  and  the  remainder  is  —  7. 

Dividing  2  by  x  —  2,  there  is  no  quotient,  and  2  itself  is  the  remainder. 


2 

-19 

59 

-60  L2 

4 

-30 

58 

2 

-15 

29 

-2 

4 

-22 

2 

-11 
4 

7 

21    -7 

Hence,  the  transformed  equation  is  2  2/'  —  7  2^2  ^  7  y  _  2. 
In  practice  the  first  detached  coefficient  is  written  but  once. 

Transform  the  following  to  equations  whose  roots  are  3  less : 

2.  a^  — 12  cc^  +  47  a;  —  60  =  0.     Prove  by  solving  both  equations. 

3.  a;<-4a^-19ar'  +  46a;  +  120  =  0. 

4.  3a;*-19a^  +  21ar^+31a;  +  12  =  0. 

5.  a;*  -  2.75  x2  +  .5  a;  +  4.5  =  0. 

675.   To  increase  the  roots  by  a  given  number. 

Since  increasing  the  roots  of  an  equation  by  h  may  be  regarded 
as  decreasing  them  by  —  h,  the  rule  for  increasing  the  roots  by  h 
is  the  same  as  that  given  for  decreasing  the  roots  by  h,  except 
that  y+h  and  x+h  take  the  place  of  y  —  h  and  x—li,  respectively. 

Examples 
Tiansform  the  following  to  equations  whose  roots  are  5  greater: 
1.    a;*+-9ar5  +  29a^  +  39a;+-18  =  0. 


THEORY  OF  EQUATIONS  511 

2.   ics  +  20  a^  +  131  aj  +  280  =  0. 

^  4.    or'^  +  loa^-f  71 CC  + 105  =  0. 

676.    To  remove  any  assigned  term  except  the  first. 

It  is  often  desirable  to  transform  an  equation  into  another 
lacking  a  certain  term.  This  is  done  by  decreasing  the  roots  by 
such  a  number  h  that  the  coefficient  of  the  assigned  term  in  the 
transformed  equation  shall  be  zero. 

Keferring  to  the  formula  for  the  equation  whose  roots  are  h 

less  than  those  of  the  given  equation,  §  674,  (3),  it  is  seen  that 

P 
the  second  term  is  lacking  when  nh+p=  0,  that  is,  when  h= ; 

that  the  third  term  is  lacking  when  h  has  such  a  value  that 
n{n-l)j^2  +  («  -  l)ph  +  g  =  0 ;  etc. 

Si 

Examples 

1.  Transform  a;^  +  6a^  +  5x— 12=0  to  an  equation  lacking 
the  second  power  of  the  unknown  number. 

Solution.  —  In  this  equation  p  =  6  and  n  =  3.  Therefore,  the  value  of  h 
tliat  makes  nA  +  j>  =  0  is  —  2  ;  that  is,  the  roots  of  the  given  equation  should 
be  decreased  by  —  2. 


6 

6 

-12L 

-2 

-2 

-8 

6 

4 

-3 

-   6 

-2 

-4 

2 

-7 

-2 

0 
Hence,  the  transformed  equation  is 

y3  4.0y2_7y_6  =  0,  or«/3-7y-6=0. 
2.    Change  207*  —  5a^  —  a;4-4  =  0  to  the  form  o^  -f-  oa;  +  ft  =  0 
with  integral  coefficients. 

Solution.  — The  reduced  form  isa^  —  ^x^  —  ^x  +  2  =  0.  (1) 

Since  A  =  —  -  =  5,  we  first  multiply  the  roots  by  6  in  order  that  h  may 
n     0 

be  an  integer. 

When  y  =  6  a;,  (1)  becomes  y*  -  15  y^  _  18  y  +  432  =  0.  *         (2) 

In  (2),  p  =  -  15  and  n  =  3.     .-.  A  =  5. 

ADV.  ALG.  —  37 


678  THEORY  OF  EQUATIONS 

Decreasing  the  roots  of  (2)  by  6  as  in  Ex.  1, 

when  z  =  y-b,  z^  -  93  ^  +  92  =  0.  (8) 

Since  j/  =  6  a;  and  2;  =  y  —  6,  «  =  6a;  —  5;  that  is,  each  root  of  (3)  is  5  ]#ss 
than  6  times  the  corresponding  root  of  (1). 

Change  to  the  form  a^  +  ax  +  &  =  0  with  integral  coefficients, 
and  express  the  roots  of  the  transformed  equation  in  terms  of 
those  of  the  given  equation; 

3.  a^-12ic2_^43a._40  =  0.        6.    a? +  3? -2  =  0. 

4.  a:3_3a;2_883,_240  =  0.        7.    x'  +  ^  +  x  +  2  =  Q. 

5.  a^-f  15a^  +  68a;  +  96  =  0.        8.    2a^-a^-2a;  +  l  =  0. 

9.  Transform  a^-|-4a^  —  7a;^  —  22a;+24  =  0  into  an  equation 
lacking  the  third  power  of  the  unknown  number.     Solve. 

677.  To  change  the  roots  to  their  reciprocals. 

Given    f{x)  =  a;"  +  pa;"-^  +  ^x"-^  -| \-ro^  +  sx+t  =  0.       (1) 

Let  -=y.     Then,  a;  =  -.     Substituting  this  value  for  x, 
X  y 

yn        yn-1         yn-i  yl        y 

Clearing  of  fractions  and  arranging  in  descending  powers  of  y, 

ty^  +  s2/»-i  +  7^-2  +  ...  +  9/  +^^2/  +  1  =0 .  (2) 

Equation  (2)  is  therefore  the  equation  whose  roots  are  the 
reciprocals  of  the  roots  of  (1).  The  transformed  equation  evi- 
dently has  the  same  coefficients  as  the  given  equation,  but  in  the 
reverse  order. 

DESCARTES'  RULE  OP  SIGNS 

678.  In  any  series  of  algebraic  numbers,  every  sequence  of  two 
unlike  signs  is  called  a  Variation  of  sign  and  every  sequence  of 
two  like  signs  is  called  a  Permanence  of  sign. 

In  ac^  _  2  X*  +  a;3  +  3  x2  —  5  a;  —  4  there  are  three  variations,   H , }-, 

-\ ,  and  two  permanences,  +  +, .    In  a;^  +  3  a;^  _  5  x  —  4,  one  variation 

and  two  permanences  occur. 


THEORY  OF  EQUATIONS  579 

679.    Let  the  signs  of  the  first  member  of  an  equation  in  x  be 

+  0 +  +  +0+-  +  +, 

in  which  0  repr.esents  the  coefficient  of  each  missing  term. 
Multiply  by  a;  —  a  and  so  introduce  a  positive  root. 

+   0 +  +  +0+-  +  + 

+  - . 

+   0 +  +  +0+-  +  4- 

-O-j-4- 0-+ 


H ±  +  ±±-  +  -  +  ±- 

Since  the  multiplicand  has  unlike  signs,  to  each  sign  of  the 
multiplicand  immediately  following  a  like  sign  there  corresponds 
an  ambiguous  sign  in  the  product.  It  will  be  observed,  also,  that 
each  ambiguous  sign  or  set  of  ambiguous  signs  in  the  product 
stands  between  two  unlike  signs,  only  interrupting  a  variation. 

Therefore,  if  the  ambiguous  signs  of  the  product  and  the  corre- 
sponding signs  of  the  multiplicand  are  stricken  out,  the  number  of 
variations  will  be  unchanged  in  the  multiplicand,  and  unchanged 
or  diminished  in  the  product. 

Hence,  the  least  possible  number  of  variations  added  by  intro- 
ducing the  positive  root  a  is  found  by  comparing  the  remaining 
signs  of  the  multiplicand  and  product. 

4-0--h0+--f 
+ +  -  +  -  +  - 

It  is  evident  that  each  variation  in  the  multiplicand  corresponds 
to  at  least  one  variation  in  the  product,  and  that  the  product  has 
in  addition  one  variation  at  the  end,  arising  from  multiplying  the 
last  term  of  the  multiplicand  by  x  and  then  by  —  a. 

Hence,  each  introduction  of  a  positive  root  adds  at  least  one  varia- 
tion of  sign. 

Let  F(x)  be  the  product  of  all  the  factors  corresponding  to  the 
negative  and  imaginary  roots  of  f(x)  =  0,  and  let  a,  b,  •-•  he  the 
positive  roots. 

Since  the  introduction  of  each  positive  root,  when  F(x)  is 
multiplied  by  (x  —  a),  (x  —  b),  •■•,  causes  an  increase  of  at  least  one 
in  the  number  of  variations  of  sign,  in  an  equation  f(x)  =  0,  the 
number  of  positive  roots  may  be  equal  to  or  less  than  the  number 
of  variations,  but  cannot  be  greater  than  this  number. 


680  THEORY   OF  EQUATIONS 

Again,  since  X—  x)=Q  has  the  roots  oif(x)  =  0  with  their  signs 
changed,  the  negati-ve  roots  of  f(x)  =  0  correspond  to  the  positive 
roots  otf(^—x)=  0.  Therefore,  the  number  of  negative  roots  of 
f(x)  =  0  may  be  equal  to,  or  less  than,  the  number  of  variations 
of  sign  in  /( — .  a;)  =  0,  but  cannot  be  greater  than  this  number. 
Hence, 

An  equation  f(x)  =  0  cannot  have  more  positive  roots  than  there 
are  variations  of  sign  in  f{x),  nor  more  negative  roots  than  there  are 
variations  of  sign  inj\—  x). 

This  is  Descartes'  rule  of  signs. 

Note.  —  In  applying  Descartes'  rule  it  will  be  understood  that  all  zero 
roots  have  been  removed.  Also  af>,  the  power  of  x  involved  in  the  absolute 
term,  will  be  regarded  as  an  even  power  of  x. 

680.  The  following  may  be  deduced  directly  from  Descartes' 
rule: 

1.  If  the  signs  of  all  the  terms  of  f{x)  are  positive,  the  equation 
f(x)  =  0  Jias  no  positive  roots. 

2.  If  all  the  terms  of  f(x)  that  involve  even  powers  of  x  have  the 
same  sign  and  all  that  involve  odd  powers  of  x  have  the  opposite 
sign,  the  equation  f(x)  =  0  has  no  negaiive  roots. 

3.  If  the  eqxiation  f{x)  =  0  is  complete,  it  cannot  have  more  nega- 
tive roots  than  there  are  permanences  of  sign  in  f{x). 

The  equation  x^  +  x*  +  a;^  +  5x  +  4  =  0  has  no  positive  roots,  for  there  are 
no  variations  of  sign  in  its  first  member. 

The  equation  a*  +  x*  —  a;3_5x  +  4=:0  has  no  negative  roots,  for  there 
are  no  variations  of  siga  in  /(  —  x) ,  or  in  x^  +  x*  +  x^  +  5  x  +  4. 

The  equation  x*  +  2x3  —  Sx^— 2x  +  2  =  0  cannot  have  more  than  2 
negative  roots,  which  is  the  number  of  permanences  in  /(x)  ;  for  since  the 

equation  is  complete,  each  permanence  of  sign  in  +  + \-,  the  signs  of 

/(x),  corresponds  to  a  variation  of  sign  in  H h  +,  the  signs  of  /(—  x). 

681.  Existence  of  imaginary  roots. 

It  is  often  possible  to  detect  the  existence  of  imaginary  roots 
by  applying  Descartes'  rule.  For,  first  removing  zero  roots,  if 
any,  if  the  sum  of  the  greatest  possible  number  of  positive  roots 
and  the  greatest  possible  number  of  negative  roots,  computed  by 
Descartes'  rule,  is  less  than  the  whole  number  of  roots,  the  number 
of  roots  remaining  must  be  the  least  possible  number  of  imaginary 
roots. 


THEORY  OF  EQUATIONS  581 

Thus,  «*  +  2a;  —  1=0  cannot  have  more  than  one  positive  root  nor  more 
than  one  negative  root.  But  the  whole  number  of  roots  is  four.  Hence, 
there  are  at  least  two  imaginary  roots. 

Examples 
Find  the  nature  of  the  roots  of  the  following : 

1.  a-*+4a^+6x2+5a;+2=0.       5.    a;*  -  4  ar'*  +  27  =  0. 

2.  x-^- 3 ar'-x +3  =  0.  6.    2(?-dx-2  =  0. 

3.  a;*-10a;2_20.T-16  =  0.       7.   ar'-l=0. 

4.  a;«  +  u;^  +  a^  +  X  =  0.  8.    a:^  +  1  =  0. 

9.  If  the  terms  of  f{x)  are  all  positive  and  each  involves  an 
even  power  of  x,  show  that  the  roots  of  the  equation  f(x)  =  0  are 
all  imaginary. 

10.  Show  that  a;"  —  1  =  0  has  two  real  roots  when  n  is  even, 
one  positive  and  one  negative,  but  only  one  real  root  when  n  is 
odd. 

MULTIPLE  ROOTS 

682.  If  an  equation  has  two  roots  each  equal  to  a,  a  is  called  a 
double  root  of  the  equation ;  if  an  equation  has  three  roots  each 
equal  to  a,  a  is  called  a  triple  root  of  the  equation ;  in  general,  if 
an  equation  has  m  roots  each  equal  to  a,  a  is  called  a  Multiple 
Root  whose  order  of  multiplicity  is  m. 

683.  Let  /(a;)  =  (a;-a)'»F(a;)  =  0  (1) 
be  an  equation  having  m  roots  each  equal  to  a. 

Then,  §  642, 

f>(x)=^(x  -  arF(x)  +  (x  -  arfF(x) 

=  m{x  —  ay-'^Fix)  +  {x  —  a)'^F'(x) 
=  (x-  a)'^-^lmF(x)  +  (x-  a)F'{x)y  (2) 

By  (2),  f'(x)  =  0  has  (m  —  1)  roots  each  equal  to  a.     Hence, 
Principle  7.  —  If  a  occurs  m  times  as  a  root  of  an  equation 
f(x)  =  0,  it  occurs  (m— 1)  times  as  a  root  of  the  first  derived  equation 
f'(x)=0. 

Thus,  X* -6a;2  + 8a;- 3  =  0,  or  (x  -  l)^(x  +  S)  =  0.  has  the  root  1 
occurring  three  times,  and  the  derived  equation  4  x^  —  12  a;  +  8  =  0,  or 
4(x  —  l)"^(x  +  2)  =  0,  has  this  root  occurring  twice. 


582 


THEORY  OF  EQUATIONS 


684.    Let  <f>{x)  denote  the  H.  C.  D.  of  f(x)  and  f{x). 

If  f(x)=0  has  no  multiple  roots,  cl>(x)  =  1 ;  but  if  /(a;)=0  has 
multiple  roots  corresponding  to  the  factors  (a;  — a)"*,  (x  —  by,  •••, 
then,  Prin.  7,  <t}{x)  =  (x—d)'^-\x  —  by-'^--. 

Hence,  if  <f)(x)  =  0  can  be  solved,  the  multiple  roots  of  f(x)  =  0 
may  be  found  and  removed. 

Examples 

1.    Solve  ic^-4a;«  +  5ar'-6a;*  +  32ar'-16a;-32  =  0. 

Solution.  —  Let    /(x)  =  a;^  -  4  a;6  +  5  a;5  _  6  x*  +  32  x'^  -  16  x  -  32. 

Then,  /'(x)  =  7  x"  -  24  x»  +  25  x*  -  24  x"  +  64  x  -  16. 

The  H.  C.  D.  of  /(x)  and  /'(x)  is  found  to  be  0(x)  =  x^  -  3  x2  +  4. 

Put  x8  -  3  x2  +  4  =  0. 

Solving  by  trial,  the  roots  of  0(x)  =  0  are  x  =  —  1,  2,  2. 

Therefore,  —  1  is  a  root  of  /'(x)  =  0  and  2  is  a  double  root  of  /'(x)  =  0. 

Hence,  Prin.  7,  —  1  is  a  double  root  and  2  is  a  triple  root  of  /(x)  =  0. 

Removing  the  corresponding  factors  (x  +  l)2(x  —  2)^  from  f(^x),  the  given 
equation  is  depressed  to  the  quadratic  equation  x^  +  4  =  0,  whose  roots  are 
2V'—  1  and  —  2V—  1.  Hence,  the  roots  of  the  given  equation  are  —  1,  —  1, 
2,  2,  2,    2V^,  -  2V^^. 

Solve  the  following  equations : 

2.  x*-x'-3ar'-^5x-2=:0. 

3.  x^  +  6x*  +  14:X^  +  20a^-\-24.x  +  16  =  0. 

4.  a^-7x'  +  Ua^-2x'-15x  +  9  =  0. 


LOCATION  OF  REAL  ROOTS 

685.  Let  f(x)  =x^ +  3x^ -llx- 5.  Since,  §  641,  f(x)  is  con- 
tinuous, if  two  numbers  substituted  for  x 
give  results  with  opposite  signs,  f(x)  passes 
through  zero  and  its  graph  crosses  the 
X-axis  once  or  an  odd  number  of  times; 
but  if  they  give  results  with  the  same 
sign,  the  graph  does  not  cross  the  X-axis 
during  the  interval  or  else  crosses  it  an 
even  number  of  times. 

For  example,  the  given  graph  does  not 
cross   the   X-axis  between    —  4   and    —  1,   but    crosses    it   once 


THEORY  OF  EQUATIONS  583 

between  —  1  and  0,  twice  between  —  1  and  3,  and  three  times 
between  —  5.2  and  3. 

But  each  intersection  gives  a  root  of  f{x)  =  0.     Hence, 

Principle  8.  —  If  two  real  numbers  substituted  for  x  in  the  first 
member  of  the  general  equation  f(x)  =  0  give  results  with  opposite 
signs,  an  odd  number  of  real  roots  of  the  equation  lie  between  the 
numbers;  but  if  they  give  results  with  the  same  sign,  no  roots  or  an 
even  number  of  roots  lie  between  the  numbers. 

Corollary  1.  —  An  equation  of  an  odd  degree  has  at  least  one 
real  root  of  a  sign  opposite  to  that  of  its  absolute  term. 

For,  if  X  =  —  00,  /(x)  is  negative  ;  if  x  =  0,  /(x)  has  the  sign  of  the  abso- 
lute term;  if  x=+oc,  /(x)  is  positive.  Hence,  if  the  absolute  term  is 
positive,  /(x)  changes  sign  for  some  value  of  x  between  —  oo  and  0  ;  but  if 
the  absolute  term  is  negative,  /(x)  changes  sign  for  some  value  of  x  between 
0  and  +  CO. 

Corollary  2.  —  An  equation  of  an  even  degree  whose  absolute 
term  is  negative  has  at  least  two  real  roots,  one  positive  and  one 
negative. 

For,  if  X  =  —  oo,  /(x)  is  positive  ;  if  x  =  0,  /(x)  is  negative  ;  if  x  =  4-  oo, 
/(x)  is  positive.  Hence,  /(x)  changes  sign  for  some  value  of  x  between 
—  CO  and  0,  and  again  for  some  value  of  x  between  0  and  +  oo. 

Corollary  3.  —  Every  equation  whose  absolute  term  is  negative 
has  at  least  one  real  root. 

686.  Limits  of  the  real  roots. 

687.  If  no  positive  real  root  of  an  equation  can  be  greater 
than  a,  then  a  is  a  Superior  Limit  of  the  positive  real  roots ;  and 
if  no  negative  real  root  can  be  less  than  —b,  then  —  6  is  an 
Inferior  Limit  of  the  negative  real  roots. 

688.  Let    f{x)=x''+px''-'^  +  qx''-^-\ [-sx  +  t  =  0  (1) 

be  an  equation  having  one  or  more  of  its  roots  positive.     Then, 
by  Descartes'  rule,  some  of  the  terms  oif{x)  iuust  be  negative. 

To  find  a  superior  limit  to  the  positive  roots,  let  r  denote  the 
number  of  positive  terms  preceding  the  first  negative  term,  miss- 
ing terms,  if  any,  being  counted ;  and  let  —  JV  be  the  negative 
coefficient  of  greatest  absolute  value. 

For  positive  values  of  x,  the  sum  of  the  terms  beginning  with 


584  THEORY  OF  EQUATIONS 

the  second  and  ending  with  the  rth  is  positive,  oi'  if  all  are  miss-' 
ing,  zero.  Hence,  dropping  these  terms  will  not  increase  f(x). 
Again,  making  —N  the  coefficient  of  each  term  after  the  rth  will 
not  increase  f(x). 

Therefore,  every  positive  value  of  x  that  makes 

a;"  —  iV(ic"-''  +  tc"-'-^  H \-x  +  l) 

positive  makes  f(x)  positive,  f{x)  having  the  same  or  a  greater 

value.  A^Ca^-'+i-n 

x»  —N(af-'  +  x"-'-^  H h  a;  + 1)  =  a;»  —  ^^^^ ±1, 

X  —  1 

when  a?  >  1,  >  aj" 

x  —  1 

Hence,  when  x>l,  f(x)  is  positive  if 

X'" =  0  or  is  positive : 

x  —  1 

that  is,  if        x" {x—1)  —  Nx'*~'"^^  =  0  or  is  positive ; 
that  is,  if  af~'^(x  —  1)  —  i\^=  0  or  is  positive. 

But  when  x>l,  a;*-^  >(x  —  ly-^  and  x"'^  (x  —  1)  >  {x  —  Vf. 
.:  of-'  {x-1)-  N>  (x  -  ly  -  N. 

Hence,  when  a;  >  1,  f{x)  is  positive  if 

(a;  —  l)*"  —  ^=  0  or  is  positive; 
that  is,  if  (x  —  ly  is  equal  to  or  greater  than  N;  that  is,  if  x  is 
equal  to  or  greater  than  ^/N^  1. 

Hence,  for  all  values  of  x  equal  to  or  greater  than  -\/N-\-l,f(x) 
is  positive. 

Since  f{x)>0  for  all  values  of  x  equal  to  or  greater  than 
■\/N+  1,  no  real  root  of  f(x)  =  0  can  equal  or  exceed  S/^+  1 ; 
that  is,  -\/N  +1  is  a  superior  limit  of  the  positive  real  roots  of 
f{x)^0. 

Thus,  in  a:*  +  «»  -  7  a;2  -  13  x  -  6  =  0,  r  =  2,  and  2V  =  13.  Therefore, 
VT3  +  1,  or  4.6  + ,  is  a  superior  limit  to  the  positive  roots. 

When  the  superior  limit  found  is  not  an  integer,  it  is  usually  convenient 
to  take  the  next  higher  integer  for  a  superior  limit.  For  example,  in  the 
illustration  just  given  6  may  be  taken  for  a  superior  limit. 

689.  To  find  an  inferior  limit  to  the  negative  roots  of  an  equa- 
tion, change  the  signs  of  all  the  roots  by  Prin.  5,  and  find  a  supe- 
rior limit  to  the  positive  roots  of  the  transformed  equation.    This 


THEORY    OF  EQUATIONS  585 

number  with  its  sign  changed  will  be  an  inferior  limit  to  the 
negative  roots  of  the  given  equation. 

Examples 
Find  the  integral  limits  to  the  real  roots  of 

1.  a:*  +  ar'  + 7  a^- 8 a;-3  =  0. 

2.  ar'-3a;3-6a;2-14a;-8  =  0. 

3.  x«  -  28 a^- 49  0-^  +  112.^  +  132  =  0. 

4.  a;«+ar^-4ic2  +  6ic-20  =  0. 

5.  a;'-a.-^  +  3a;''  +  76a;3-243ar'  +  108  =  0. 

690.  The  utility  of  the  preceding  principles  and  their  failure 
in  certain  cases  may  be  illustrated  in  the  solution  of  such  an 
equation  asa^  —  ar^  —  4ar  +  3a;  +  3  =  0. 

Since  no  factor  of  the  absolute  term  satisfies  the  equation, 
Prin.  Ij  Cor.  2,  the  equation  has  no  integral  root.  Also,  Prin.  3, 
the  equation  has  no  rational  fraction  for  a  root.  Hence,  the 
equation  has  no  commensurable  root. 

If  the  equation  has  a  positive  root,  removing  it  would  leave  a 
depressed  equation  of  an  odd  degree  with  a  negative  absolute 
terra.  Since,  Prin.  8,  Cor.  1,  such  an  equation  has  a  positive  root, 
if  the  given  equation  has  one  positive  root,  it  has  two.  Similarly, 
if  the  given  equation  has  one  negative  root,  it  has  two.  But  we 
cannot  tell,  without  actual  trial,  whether  the  number  of  positive 
roots  or  of  negative  roots  is  zero  or  two. 

By  §  688,  if  the  equation  has  real  roots,  their  superior  limit  is 
5,  and  their  inferior  limit  is  —  3.  " 

Substituting  -  2,   - 1,  0,  1,  2,     3,       4,  for  x, 
we  obtain  5,  -  2,  3,  2,  1,  30,  143,  for  f(x). 

By  Prin.  8,  then,  a  negative  root  lies  between  —  2  and  —  1,  and 
another  between  —  1  and  0.  But  without  considerable  labor  in 
substituting  values  between  0  and  4,  it  cannot  be  determined 
whether  the  other  two  roots  are  positive  or  imaginary. 

When  two  roots  of  an  equation  are  nearly  equal  or  when  there 
are  imaginary  roots,  the  real  roots  are  not  found  readily  by  trial. 
In  such  cases  the  number  of  real  and  imaginary  roots  and  the 
situation  of  the  real  roots  may  be  completely  determined  by 
Sturm's  theorem,  which  follows. 


686 


THEORY  OF  EQUATIONS 


691.   Sturm's  theorem. 

Let  f{x)  =  0  be  an  equation  in  x  freed  of  multiple  roots. 

Then,  f{x)  and  f\x)  have  no  common  factor  involving  x. 

Divide  J{x)  by  /'(x),  as  in  the  process  of  finding  the  H.C.D., 
until  the  remainder  is  of  lower  degree  than  the  divisor,  introduc- 
ing or  rejecting  positive  monomial  factors,  if  desired,  hut  not  nega- 
tive ones.  Denote  the  remainder  with  its  sign  changed  by  F^jix). 
Similarly,  divide  f'{x)  by  F^ix)  and  denote  the  remainder  with  its 
sign  changed  by  -^3(0;).  Continuing  in  this  way,  since  f(x)  and 
f'{x)  have  no  common  factor  involving  x,  a  remainder  not  involv- 
ing x  will  finally  be  obtained.  Denote  this  remainder  with  its  sign 
changed  by  F^(x).  Then,  J{x),  f<{x),  F^(x),  F,(x),  ■  ■  -,  F^_^(x),  fJx) 
are  called  Sturm's  functions. 
Thus,  let  f(x)  =  a;3  +  a.-2  -  2  X  -  8,  whence  /'(a;)  =  3  x2  +  2 x  -  2. 
3  +  2- 


-13 


3  +  15 


-13-2 
-  13  -  66 


+  63 


1  +  1- 

2- 

8 

3 

. 

3  +  3- 

U- 

24 

3  +  2- 

2 

1  - 

4- 

24 

3 

3- 

12- 

72 

3  + 

2- 

2 

+  14)- 

14- 

70 

— 

1  - 

5 

1  + 

5 

F^ix)  =  X  +  6. 


i?'3(x)=-63. 


Hence,  Sturm's  functions  are  x3+  x^  —  2  x  —  8,  3  x^  +  2  x  —  2,  x  +  5,  —  63. 

Sturm's  theorem  may  be  stated  as  follows : 

If  a  and  b  are  two  real  numbers,  b  greater  than  a,  then  the  number 
of  variations  of  sign  obtained  by  writing  f (a),  f  (a),  F2(a),  •••,  F„(a) 
in  order,  less  the  number  of  variations  obtained  by  writing  f(b),  f'(b), 
Fi(b),  •••,  FJb)  in  order,  is  equal  to  the  number  of  real  roots  of 
f{x)  =  0  that  lie  betioeen  a  and  b.  , 

The  proof  is  as  follows : 

1.  Since  fix)  and  f'{x)  have  no  common  factors,  by  §  148  no 
two  successive  functions  have  a  common  factor;  that  is,  no  two 
successive  functions  become  zero  for  the  same  value  of  x. 

Let  Fr{x)  be  any  of  Stm-m's  functions  except  f(x)  and  Fn(x). 


THEORY  OF  EQUATIONS  587 

By  the  definition  of  Sturm's  functions,  if  i^r-i(^)  is  divided  by 
Fr{x),  tlie  remainder  is  —  F^^^{x).     Represent  the  quotient  by  Q. 

Then,  §  102,         2^,_i(^)  =  QF^x)  -  F^^lx) 
\iF,{x)=0,  =-F,.^,(x); 

that  is,  if  any  intermediate  fundioyi  becomes  zero  for  any  value  of  x, 
the  aOjax^ent  functions  have  opposite  signs  for  that  value  of  x. 

Let  c'  be  a  root  of  F^x)  =  0,  so  that  F^{c')  —  0;  and  let  c'  —  h 
and  c'  +  /t  be  so  close  to  c'  that  no  roots  of  the  adjacent  functions 
lie  between  c'  —  h  and  c'  +  h. 

Then,  the  adjacent  functions  have  opposite  signs  for  x  =c'  and 
keep  these  signs  while  x  increases  from  c'  —  ^  to  c'  +  h. 

Since  both  before  and  after  x  =  c'  the  sign  of  F^x)  is  like  that 
of  one  of  the  adjacent  functions  and  unlike  that  of  the  other, 
ichen  X  is  passing  through  a  root  of  any  intermediate  function, 
Sturm's  functions  maintain  the  same  number  of  variations. 

2.  Let  c  be  a  root  of  f(x)  =  0,  so  that  /(c)  =  0,  and  let  h  be  a 
positive  number,  as  small  as  we  please. 

Then,  /(c  —  h)  denotes  the  value  of  f{x)  just  before,  and  /(c  +  h) 
denotes  the  value  of  f{x)  just  after,  x  passes  through  the  root  c. 

§  640,  /f(c  -  h)  =  f{c)-  hf\c)  +,|V"(c)-  -, 

and  /(c  +  h)  =  f{c)+  hf'(c)  +~f"ic)+  .... 

Since  f(c)  =  0,  if  h  is  taken  sufficiently  small  the  signs  of 
/(c  —  h)  and  f(c  +  h)  will  be  the  same  as  those  of  their  numeri- 
cally greatest  terms,  —  hf'(c)  and  hf'(c),  respectively.  Then,  h 
being  positive,  f(c  —  h)  has  the  sign  opposite  to  that  of /'(c),  and 
/(c  -f  h)  has  the  same  sign  as  f'(c). 

That  is,  as  x  increases  through  a  root  of  f(x)  =  0,  f[^x)  changes 
from  a  number  having  the  sign  opposite  to  that  of  f'(x)  to  a 
number  having  the  same  sign  as  f'{x).  Hence,  as  x  increases 
thi'ongh  a  root  off(x)  —  0,  Sturm's  functions  lose  one  variation. 

Therefore,  since,  as  jx  increases  continuously  from  a  to  b, 
Sturm's  functions  lose  one  variation  every  time  x  passes  through 
a  root  of  ^f{x)  =  0,  but  lose  none  when  x  passes  through  the  roots 
of  the  other  functions,  the  number  of  variations  lost  between  a  and 
b  indicates  the  number  of  real  roots  between  a  and  b. 


588  THEORY  OF  EQUATIONS 

692.  Since  all  the  negative  roots  of  an  equation  lie  bet\ve»-.i 
—  00  and  0,  if  —  00  is  substituted  for  x  in  Sturm's  functions  and 
then  0  is  substituted,  the  number  of  variations  lost  gives  the 
number  of  negative  roots  of  the  equation. 

Similarly,  if  0  is  substituted  for  x  in  Sturm's  functions  and 
then  4-  00  is  substituted,  the  number  of  variations  lost  gives  the 
number  of  positive  roots  of  the  equation. 

Examples 

1.    Locate  the  roots  of  x-*  —  x^  —  4  a;^  +  3  a;  -f-  o  =  0. 

Solution.  —  In  §  OiH)  it  was  found  that  one  negative  root  of  this  equation 
was  situated  between  —  2  and  —  1  and  another  between  —  1  and  0.  Tlie 
nature  and  situation  of  the  other  two  roots  are  found  as  follows  : 

Let  f{x)  =  X*  -  x3  -  4  x2  +  3  X  +  3. 

Then,  /'(x)  =  4  x^  -  3  x^  -  8  x  +  3. 

Employing  the  process  of  finding  the  H.C.I),  of /(x)  and /' (x) ,  taking  care 
not  to  introduce  or  reject  negative  factors,  it  is  discovered  that/(x)  and/'(x) 
have  no  common  factor.  Hence,  /(x)  =  0  has  no  multiple  roots,  and  /(«) 
and  /'(x)  are  the  first  two  of  Sturm's  functions. 

Representing  the  successive  remainders  with  their  signs  changed  by  i^2(x), 
i?3(x),  and  Fi(x),  Sturm's  functions  are: 

/(x)  =  X*  -  x3  -  4  X--2  +  3  X  +  3 
/'(x)  =  4x''-3x2-8x  +  3 
i?'2(x)  =  35x2-28x-51 
Fsix)  =  11  X  -  18 
Fiix)  =  + 

Substituting  —  co,  0,  and  +  oo  for  x  in  these  functions, 


X 

Kx) 

/'(x) 

F,(x) 

Fs(x) 

F^ix) 

variations 

—  00 

+ 

- 

+ 

— 

+ 

4 

0 

+ 

+ 

- 

- 

+ 

2 

+  00 

+ 

+ 

+ 

+ 

+ 

0 

Since  two  variations  are  lost  between  —  oo  and  0,  by  Sturm's  theorem 
there  are  two  negative  roots,  as  before  ascertained  ;  and  since  two  variations 
are  lost  between  0  and  +  co,  the  other  two  roots  are  real  and  positive. 

By  §  688,  5  is  a  superior  limit  of  the  positive  roots. 

Substitute  in  Sturm's  functions  0,  1,  2,  3,  •••,  or  as  many  of  these  numb«>i^« 
as  may  be  necessary  to  lose  two  variations. 


THEORY 

OF  EQUATIONS 

58£ 

z 

/(x) 

f'C^) 

Fi(x) 

Fs(^) 

F,(x) 

variations 

0 

+ 

+ 

- 

— 

+ 

2 

1 

+ 

- 

- 

- 

+ 

2 

2 

+ 

+ 

+ 

+ 

+ 

0 

It  is  discovered  tliat  two  variations  are  lost  between  1  and  2 ;  therefore, 
there  are  two  positive  roots  between  1  and  2. 

Locate  the  real  roots  of  the  following  equations  and  find  the 
number  of  imaginary  roots  : 

2.  2ar^-15x'2  4-32a;-21  =  0. 

3.  16ar'-4x-2-80a;  +  75  =  0. 

4.  a:^-2ar^-7ar'  +  10x  +  10  =  0. 

5.  a;*-x-2_l0x-4  =  0. 

693.    Homer's  method  of  solving  numerical  equations. 

Commensurable  roots  expressed  by  two  or  more  figures  may  be 
found  exactly  by  Horner's  method,  and  real  incommensurable  roots 
may  be  found  to  any  degree  of  accuracy  required. 

1.  Let  it  be  required  to  find  approximately  the  one  positive 
real  root  of  the  equation 

f(x)=x'  +  3x'-nx-5  =  0.  (1) 

Substituting  0,  1,  2,  3,  •••  for  x,  the  corresponding  values  of  the 
first  member  are  /(O)  =  -  5,  /(I)  =  - 12,  /(2)  =  -  7,  /(3)  =  16,  •  •  -. 

Since  f(x)  is  negative  when  x  =  2  and  positive  when  x  =  3,  the 
one  positive  root  must  lie  between  2  and  3  (Prin.  8). 

Transforming  (1)  to  an  equation  whose  roots  are  2  less, 


3 

-11 

-5 

2 

10 

-2 

5 

-    1 

-7 

2 

14 

7 

13 

2 

9 


tho  first  transformed  equation  is 

f  +  9f  +  13y-7 


(2) 


690 


THEORY   OF   EQUATIONS 


Since  y  =  x  —  2  and  x  lies  between  2  and  3,  the  corresponding 
value  of  y  lies  between  0  and  1.     Try  .5. 

19  13  -7  [^ 

.5  4.75  8.875 

9^5         17.75     !         1.875 
Since  ^-f92/^  +  13y  —  7  is  equal  to  1.875  when  y  =  .5  and  is 
equal  to  —  7  when  y  =  0,  the  root  of  this  equation  must  be  less 
than  .5.     Trying  .4,  it  is  found  that  the  function  does  not  change 
sign.     Hence,  the  root  lies  between  .4  and  .5. 

Transforming  (2)  to  an  equation  whose  roots  are  .4  less, 


9 

13 

.4 

3.76 

9.4 

16.76 

.4 

3.92 

9.8 

1  20.68 

.4 

-7 
6.704 


Li 


.296 


^ 


.^1 


10.2 


the  second  transformed  equation  is 

z«  +  10.2  z^  +  20.68  z  -  .296  =  0,  (3) 

in  which  z  =  y  —  A  =  (x  —  2)  —  A  =  x  —  2.4. 

Since  the  positive  root  of  (2)  lies  between  .4  and  .5  and  each 
root  of  (3)  is  .4  less  than  the  corresponding  root  of  (2),  the 
positive  root  of  (3)  lies  between  0  and  .1.  Since  this  value  of  z 
is  a  small  fraction,  in  (3)  the  higher  powers  of  z  are  small  as 
compared  with  the  first  power.  Hence,  the  positive  root  of  (3)  is 
very  nearly  the  same  as  the  root  of 

20.68  z  -  .296  =  0,  or  20.68  z  =  .296. 

Therefore,  z  =  .01  +,  and  the  next  figure  of  the  root  is  1. 
Transforming  (3)  to  an  equation  whose  roots  are  .01  less, 

1.01 


10.2 

20.68 

-.296 

.01 

-  .1021 

.207821 

10.21 

20.7821 

-  088179 

.01 

.1022 

10.22 

20.8843 

.01 

lO-J^? 


THEORY   OF  EQUATIONS 


591 


the  third  transformed  equation  is 

1^  +  10^23  v"  +  20^8843  v  -  ^088179  =  0, 
in  which  v  =  z  —  .01  =  x—  2.41. 


(4) 


The  next  figure  of  the  root  is  found  by  neglecting  the  higher 
powers  of  v  in  (4)  and  solving  the  equation 

20.8843  V  -  .088179  =  0, 

which  gives  ^  =  .004+.     Hence,  the  root  is  2.414 +. 

The  preceding  transformations  and  the  work  of  substituting 
.004  for  V  in  (4)  to  test  the  last  figure  may  be  written  in  one 
process,  as  shown  below.  The  coefficients  of  the  successive  trans- 
formed equations  are  in  heavy-face  type. 


f 


^ 


3                -11 
2                    10 

-  5                 1  2.414 

-2 

5 
2 

-1 

14 

-7 
6.704 

7 
2 

13 

3.76 

-.296 

.207821 

9 

.4 

16.76 
3.92 

-  .088179 

.083700944 

9.4 
.4 

9.8 
.4 

20.68 

.1021 

20.7821 

.1022 

-  .004478056 

10.2 

.01 

10.21 

■01 

10.22 
.01 

20.8843 

.040936 
20.925236 

10.23 

.00 

4 

10.234 

The  broken  lines  mark  the  close  of  each  transformation. 
If  more  figures  of  the  root  are  desired,  the  last  transformation 
may  be  completed  and  the  process  continued  to  any  number  of 


592 


THEORY  OF  EQUATIONS 


figures  of  the  root  by  successive  transformations  in  each  of  which 
the  root  is  decreased  by  tlie  largest  number  that  will  not  cause  the 
absolute  term  to  change  sign. 

The  solution  of  an  equation  f{x)  =  0,  as  for  example 


ic3  +  3a^-llx-5  =  0, 


(1) 


whose  positive  root  has  just  been  found  to  three  decimal  places, 
is  often  facilitated  by  plotting  the  graph  oif{x). 

Plotting  corresponding  values  of  x  and  f{x)  in  (1),  the  graph 
given  on  page  582  is  obtained.  It  shows  that  the  equation  has 
two  negative  roots,  and  a  positive  root  lying  between  2  and  3. 

The  steps  in  Horner's  process  of  approximation  to  the  positive 
root  2.414+  are  illustrated  graphically  as  follows : 


In  the  accompanying  figure  the  positive  root  is  represented 
graphically  by  the  abscissa  OP.  Now  if  the  F-axis  is  moved 
parallel  to  itself  2  units  toward  the  right  to  the  position  of  0'  Y', 
the  abscissa  of  every  point  in  the  graph,  including  that  of  P,  will  be 
decreased  by  2.     For  example,  the  abscissa  of  P  will  become  O'P. 

Hence,  moving  the  Y-axis  2  units  toward  the  right  has  the  effect 
of  decreasing  the  roots  of  the  given  equation  by  2. 


THEORY  OF  EQUATIONS  593 

The  new  axes,  then,  are  O'X  and  0'  Y',  the  new  equation,  that 
is,  i\ie  first  transformed  equation  is 

2/'  +  92/^  +  132/-7  =  0,  (2) 

in  which  y  =  x  —  2,  and  the  curved  line  PiP^Pz  •  •  •  Pg,  called  the 
graph  of  a:^  4-  3  a*-  —  11  cc  —  5  with  reference  to  the  old  axes,  be- 
comes the  graph  of  i/*  +  9  ?/^  + 13  y  —  7  referred  to  the  new  axes. 

Since  y  =  x  —  2,  equation  (2)  has  a  root  between  0  and  1. 
Accordingly,  divide  the  portion  of  the  X-axis  between  the  second 
and  third  divisions  into  tenths  and  plot  the  points  Pi,  P^,  P^,  P^, 
Pg,  corresponding  to  y  —  .1,  .2,  .3,  .4,  .5,  respectively.  It  is  found 
that  between  .4  and  .5  the  ordinates  change  from  negative  to  posi- 
tive. Therefore,  the  graph  crosses  the  X-axis  between  y=  A  and 
y  =  .5,  that  is,  between  x  =  2.4  and  x  =  2.5. 

Hence,  the  F-axis  may  be  moved  .4  of  a  unit  nearer  to  P  than 
has  been  done,  without  causing  the  origin  to  pass  beyond  P. 
Making  this  transfer,  then,  the  new  axes  are  0"Xand  0"Y",  and 
the  second  transformed  equation  is 

z"  + 10.2  z^  -f  20.68  z  -  .296  =  0,  (3) 

in  which  z  =  y  —  A  =  x  —  2.4. 

Since  z  =  y  —  A,  equation  (3)  has  a  root  between  0  and  .10. 

Dividing  the  portion  of  the  X-axis  between  the  divisions  2.4 
and  2.5  into  ten  equal  parts,  each  .01  of  a  unit  in  length,  it  is 
found  by  substituting  .01,  .02,  •••  for  z  in  (3)  that  the  first  mem- 
ber changes  sign  between  z  =  .01  and  z  =  .02,  that  is,  the  graph 
crosses  the  X-axis  between  x  =  2.41  and  x  =  2.42. 

Hence,  the  F-axis  may  be  moved  .01  of  a  unit  nearer  to  P, 
without  causing  the  origin  to  pass  beyond  P.  This  decreases  the 
roots  of  (3)  by  .01  and  gives  the  third  transformed  equation 

v"  4- 10.23^^-1- 20.8843  V -.088179  =  0,  (4) 

in  which  v  =  z  —  .01  =  x  —  2.41. 

This  process  of  approximation  can  be  continued  indefinitely. 

Graphically  considered,  then,  Horner's  method  of  approximation 
consists  in  moving  the  F-axis  toward  the  right,  first,  the  greatest 
number  of  units  possible  without  causing  the  origin  to  pass 
beyond  the  intersection  of  the  graph  and  the  X-axis,  then  the 

ADV.   ALG.  —  38 


594 


THEORY  OF  EQUATIONS 


1 

;ible  1 


greatest  possible  number  of  tenths,  then  the  greatest  possible 
number  of  hundredths,  etc.  The  approximation  to  the  root  at 
any  stage  is  the  total  distance  the  origin  has  been  moved. 

2.  The  process  on  page  591  ilhistrates  the  general  method  of 
extending  a  positive  root  to  any  number  of  figures  after  finding 
enough  figures  to  separate  it  from  other  roots,  if  any,  that  are 
nearly  equal  to  the  root  sought. 

The  following  process  illustrates  certain  modifications  of 
Homer's  method  for  roots  that  are  nearly  equal,  as  explained 


\5A 


aelow. 

Let  a* 

-12ar^  +  38 

,af 

+  8 

a;  -  112  = 

0. 

1 

-12                 38 

8                -112 

6             -35 

-7                   3 

16 

115 

23 

3 

6             -10 

-35 

-  2.9824 

-2               -7 

-12 

1          .0176 

5 

15 

4.544 
-  7.456 

3 

8 

5 

3.36 

5.952 

y" 

8 

11.36 

-1.504 

.4                3.52 

8.4               14.88 

.4 

3.68 

8.8 

18.56 

.4 

9.2 
.4 

2*         9.6 

Explanation.  —  By  Sturm's  theorem  it  is  found  that  the  equation  has  a 
negative  root  and  three  positive  roots,  one  between  2  and  3,  and  two  between 
5  and  6.     The  above  process  is  concerned  with  the  roots  between  5  and  6. 

Decreasing  the  roots  by  5,  the  first  transformed  equation  is 

yi  +  8y^  +  Sy'--l2y  +  S  =  0,  (1) 

which  has  only  two  positive  roots,  both  between  0  and  1,  the  least  positive 
root  having  been  passed,  that  is,  changed  to  a  negative  root.  This  is  indi- 
cated by  the  change  of  sign  of  the  absolute  term  from  —  to  4-  and  the  loss 
of  one  variation  of  sign. 

Since  the  first  transformed  equation  has  ticn  roots  between  0  and  1,  the 
rule  that  the  next  figure  of  the  root  is  the  greatest  number  of  tenths  by 


THEORY  OF  EQUATIONS  595 

which  the  roots  can  be  decreased  without  changing  the  sign  of  the  absolute 
term  is  not  sufficient  to  guard  against  passing  the  roots,  for  it  is  possible,  by 
using  too  large  a  number,  to  pass  both  roots  at  the  same  time  and  still  leave 
the  absolute  term  positive  in  consequence  of  tioo  changes  in  sign. 

The  roots  may  be  separated,  however,  by  applying  the  rule  that  the  equa- 
tion loses  one  variation  of  sign  for  each  positive  root  that  is  passed  when 
the  roots  are  decreased  by  a  number  greater  than  the  root  in  question. 

If  the  roots  of  (1)  are  decreased  by  .4,  as  shown  in  the  process,  or  by  any 
less  number  of  tenths,  the  absolute  term  remains  positive,  and  besides,  no 
variations  are  lost.  Hence,  both  positive  roots  are  greater  than  .4.  But  if 
the  roots  are  decreased  by  .5  or  by  any  greater  number  of  tenths,  two  varia- 
tions are  lost.     Hence,  both  positive  roots  are  less  than  .5. 

That  is,  both  positive  roots  of  (1)  lie  between  .4  and  .5,  and  the  corre- 
sponding roots  of  the  given  equation  lie  between  5.4  and  5.5. 

The  second  transformed  equation 

z*  +  9.6  z^  +  18.56  22  _  1.504  z  +  .0176  =  0  (2) 

has  two  roots  between  0  and  .10.  Trying  .01,  .02,  .03,  •••,  it  is  found  that 
the  absolute  term  changes  sign  and  one  variation  is  lost  in  passing  from  .01 
to  .02,  and  again  in  passing  from  .06  to  .07.  Hence,  the  roots  of  (2)  are 
.01+  and  .06  +  ,  and  the  corresponding  roots  of  the  given  equation  are  5.41  + 
and  5.46  +  . 

The  roots  are  now  separated  and  the  above  process  may  be  continued  to 
find  either  of  them  by  the  easier  method  given  in  1. 

In  finding  the  less  root  5.414  + ,  the  absolute  terms  will  continue  to  be 
positive  ;  in  finding  the  greater  root  5.464  +,  they  will  be  negative  beginning 
with  the  third  transformed  equation. 

Rule  for  Positive  Roots. — Find  the  first  figure  of  the  root  by 
trial  or  by  Sturm's  theorem,  and  decrease  the  roots  by  this  number. 

If  the  root  is  separated  from  other  roots,  find  each  succeeding 
figure  of  the  root  by  decreasing  the  roots  by  the  greatest  number  of  the 
next  loioer  order  that  will  not  cause  the  absolute  term  to  change  sign. 

If  the  root  is  not  seiJarated  from  other  roots,  first  separate  the 
roots  by  Sturm's  theorem ;  or  decrease  the  roots  by  the  greatest 
number  of  the  next  lower  order  that  will  not  cause  the  equation  to 
lose  more  variations  than  there  are  positive  roots  that  are  less  than 
the  root  sought. 

After  several  figures  of  the  root  have  been  found,  the  next  figure  may  be 
found  u.'^ally  by  neglecting  powers  of  the  unknown  number  higher  than  the 
first  and  solving  the  resulting  simple  equation  ;  or,  if  the  coefficient  of  the 
first  power  is  small  in  comparison  with  the  coefficients  of  the  higher  powers 
or  is  equal  to  zero,  by  neglecting  the  powers  higher  than  the  second  and 
solving  the  resulting  quadratic  equation. 

ADV.  ALG.  —  38 


596 


THEORY  OF  EQUATIONS 


694.  To  avoid  decimals  and  to  abbreviate  the  work  when  many 
figures  of  the  root  are  required,  Horner's  process  is  modified  as 
explained  below.     Compare  with  the  process  on  page  591. 


f 


^ 


v^ 


3         -11 
2          10 

— 

5      12.41421 

2 

6 

2 

-1 

14 

— 

7000 

6704 

7 
2 

1300 

376 

-  296000 

207821 

90 

4 

1676 
392 

-  88179000 

83700944 

94 
4 

206800 

1021 

-  4478O5600f) 

4193648 

98 
4 

207821 
1022 

-  2844080^ 

209704 

1020 

1 

1021 

1 

20884300 

40936 

20925236 

40952 

-  74704(ji^ 

1022 
1 

2096618/^ft)() 

205 

10230 

4 

2096824 
205 

10234 
4 

10238 
4 

209702^p(f) 
1 

209704 
1 

i 

20970^|!)p 

ijD;80 

Explanation.  — 1.  Decimals  are  avoided  by  multiplying  the  roots  of  each 
transformed  equation  by  10,  the  numbers  substituted  for  y^  z,  v,  •••  being 
regarded  as  4,  1,  4,  •••  instead  of  .4,  .01,  .004,  •••. 

2.  After  several  figures  of  the  root  have  been  obtained  (in  this  case 
beginning  with  the  fourth  transformed  equation),  nearly  as  many  more 
are  accurately  found  by  a  contracted  process.  For  since  dividing  both  mem- 
bers of  an  equation  by  the  same  known  number  does  not  change  the  roots, 


THEORY  OF  EQUATIONS  597 

after  the  roots  of  the  fourth  transformed  equation  have  been  multiplied  by 
10,  we  may  divide  each  coefficient  by  1000 ;  and  if  the  decimal  parts  thus 
formed  are  cut  off,  the  accuracy  of  the  root  will  not  be  affected  for  several 
figures.  Some  allowance  for  the  decimal  cut  off  is  made,  when  convenient. 
Thus,  102.420x2  is  nearer  205  than  204  and  is  taken  as  205. 

In  practice  the  two  operations  of  multiplying  the  roots  by  10  and  dividing 
the  equation  by  1000  are  performed  simultaneously  by  omitting  to  annex 
the  ciphers  and  then  cutting  off  the  last  figure  of  the  coefficient  of  the  first 
power,  the  last  two  figures  of  the  coefficient  of  the  second  power,  and  so  on. 
In  the  above  process  this  practically  destroys  the  leading  coefficient  and  re- 
duces the  equation  to  the  quadratic  102  vfl  +  2096619  w  —  4478056  =  0,  next  to 
the  quadratic  u'^  +  209703  u  —  284408  =  0,  and  finally  to  the  simple  equation 

20970  t  -  74704  =  0. 

Solving  this  equation  by  dividing  74704  by  20970,  t  =  3.562+. 
Hence,  the  root  of  the  given  equation  is  2.4142135624- . 

695.  To  find  the  value  of  a  negative  root,  Horner's  method 
may  be  applied  to  find  the  corresponding  positive  root  of  that 
transformed  equation  whose  roots  are  numerically  equal  to  the 
roots  of  the  given  equation,  but  opposite  in  sign.  The  resulting 
positive  root  with  its  sign  changed  will  be  the  required  negative 
root. 

696.  Horner's  method  may  be  used  to  find  the  principal  nth. 
root  of  any  number.  For  example,  the  principal  fifth  root  of  2 
may  be  found  by  solving  the  equation  o^  —  2  =  0  by  Horner's 
method. 

Examples 

1.  Find  by  Horner's  method  the  root  lying  between  1  and  2  of 
the  equation  4  ic^  —  3  a^  +  .r  —  14  =  0.  Test  the  result  by  multi- 
plying the  roots  by  4  and  substituting  4  times  the  root  found. 
Find  the  imaginary  roots. 

2.  The  equation  8  ar'  —  29  cc^  -f  29  a;  —  21  =  0  has  a  root  between 
2  and  3.     Find  all  the  roots. 

3.  The  number  and  situation  of  the  real  roots  of  the  equation 
x''  —  a^  —  4a^-f3a;  +  3  =  0  have  been  discussed  in  example  1, 
page  588.     Find  all  the  roots  to  the  third  decimal  place. 

4.  Find  all  the  roots  of  .r*  -  6 ar^  -f  5 a;^  -f-  14a;  —  4  =  0  to  the 
nearest  third  decimal  place. 

5.  Find  the  fifth  root  of  330383.69407  by  Horner's  method. 


698  THEORY  OF  EQUATIONS 

697.  Newton's  method  of  approximation. 

If  one  and  only  one  real  root  of  the  equation 

f{x)  =  a;"  +pa;''-^  +  qx''-^  -\ \-sx  +  t  =  0  (1) 

lies  between  a  and  b,  and  if  a  is  less  than  this  root  by  a  number  h 
that  is  small  in  comparison  with  a,  then  a  +  /i  is  the  root,  and  in 

(a  +  hy  +  p{a  +  hy-'^  +  q{a  +  /i)"-^ -| \- s{a-^h)  +  t  =  0    (2) 

the  terms  involving  h"^,  h%  h*,  •••  may  be  neglected. 
Then,  approximately, 

a"     +pa^-^  +qa"-^  + \-  sa  +  t 

+  [wa»-i  +  p{n  —  l)a"-2  +  q{n  —  2)a"-3  H h  s]^  =  0; 

that  is,  §639,     /(a)+/'(a)fe  =  0,  or^  =  -^.  (3) 

Thus,  to  approximate  to  the  root  of  x^  —  2x— 5  =  0  lying  between  2 
and  3,  denote  the  root  by  2  +  ^  as  in  (2),  neglect  the  terms  involving  h^ 
and  h?  in   (2  +  hY  —  2(2  +  A)  —  5  =  0,   and  solve  the  resulting  equation 
lOA  —  1  =  0.     The  result  is  A  =  .1,  approximately. 
Or  proceed  as  follows,  using  the  result  given  in  (3)  : 
f{x)=x^  -  2x  -  5  and/''(a;)=:  3x2  -  2. 
.-.  /(2)  =  8  -  4  -  5  =  -  1  and  /'(2)  =  12  -  2  =  10. 
Hence,  by  (3),  h  =  —  .1^  =  .1,  and  the  root  is  2.1,  approximately. 
Next,  using  2.1  for  a,  it  is  found  that  /(2.1)=  .061  and  /'(2.1)=  11.23. 

Hence,  by  (3),  the  next  addition  to  the  root  is  —  ,  or  —  .006,  and  the 

^  ^  11.23 

root  is  2.1  —  .005,  or  2.095,  approximately. 

Note.  — This  method  possesses  but  little  practical  value. 

RECIPROCAL  BQUATIOITS 

698.  When  each  root  of  an  equation  is  the  reciprocal  of  some 
other  root  or  of  itself,  the  equation  is  called  a  Reciprocal,  or  Recur- 
ring, Equation. 

699.  Since  the  only  numbers  that  are  reciprocals  of  themselves 
are  1  and  —  1,  there  are  four  kinds  of  reciprocal  equations : 

1.  Those  whose  roots  in  pairs  are  reciprocals  of  each  other. 

2.  The  first  kind  with  the  root  1  in  addition. 

3.  The  first  kind  with  the  root  —  1  in  addition. 

4.  The  first  kind  with  the  roots  1  and  —  1  in  addition. 


THEORY  OF  EQUATIONS  699 

The  four  kinds  may  be  illustrated,  respectively,  as  follows: 

1.  6xt +  5a;3-38x=^  +  5x+6  =  0  ;  roots,  2,  |.-3,  -\. 

2.  2x3  -  7  x'^  +  7 a:  -  2  =  0  ;  roots,  2,  \,  1. 

3.  4x3-13x2-  13x  +  4  =  0;  roots,  4,  |,  -1. 

4.  4  X*  -  17  x3  +  17  X  -  4  =  0  ;  roots,  4,  \,  1,  -  1. 

It  is  evident  that  equations  of  the  second,  third,  and  fourth 
kinds  may  be  depressed,  by  dividing  by  a?  —  1,  a;  +  1,  and  x^  —  1, 
respectively,  to  equations  of  the  first  kind. 

Hence,  the  first  kind  is  the  standard  reciprocal  equation. 

700.  Let    f{x) = mx" + pa;""^ + ga;«-2 h \-r2i?-\-sx+t=0        (1) 

be  a  reciprocal  equation. 

Then,  for  all  values  of  x  that  make  f{x)  =  0, 

^-\  (2) 

Substituting  this  value  of  x  in  (1),  multiplying  both  members 
of  the  resulting  equation  by  a;",  reversing  the  order  of  the  terms, 
and  changing  signs  throughout  in  case  t  is  negative, 

±  te"  ±  sa;"-^  ±  rar"--  ±  ••  •  ±  ga;^  ±  pa;  ±  m  =  0,  (3) 

the  upper  signs  being  used  when  t  is  positive  and  the  lower  ones 
when  t  is  negative. 

Since  (1)  and  (3)  have  the  same  roots,  their  first  members  are 
composed  of  the  same  factors  and,  consequently,  are  identical. 

.-.  m=zt,  p  =  s,  q=.r,-'-;  or  else  m  =  —  t,  p  =  —s,  q  =  —r,  •••. 

Hence,  in  a  reciprocal  equation,  the  coefficients  of  the  terms  equally 
distant  from  the  first  and  last  terms  of  f(x)  are  equal  when  the  last 
term  is  positive,  and  numerically  equal  with  opposite  signs  when  the 
la^t  term  is  negative. 

701.  Let     f{x)  =  ±  ^a;"  ±  sof-'  ±  ra;"-='  + \-ra?  +  sx  +  t  =  0 

be  a  reciprocal  equation,  the   lower   signs   corresponding   to   a 
negative  absolute  term. 

If  n  is  odd,  f(x)  has  an  even  number  of  terms  that  may  be 
grouped  in  pairs  having  equal  or  numerically  -equal  coefficients. 
If  the  corresponding  terms  have  like  signs,  by  §  111,  4,  each  of 
the  groups  t(x"  +  1),  sx(x^~^  + 1),  rx^{x^~*  +  1)>  *"  contains  a;  + 1 


600  THEORY  OF  EQUATIONS 

as  a  factor,  whence  —  1  is  a  root  of  the  equation  and  the  equation 
is  of  the  third  kind;  but  if  the  corresponding  terms  have  unlike 
signs,  by  §  111,  1,  a;  —  1  is  a  factor  of  each  group,  1  is  a  root  of 
the  equation,  and  the  equation  is  of  the  second  kind. 

If  n  is  even,  f(x)  has  an  odd  number  of  terms  and  there  is  a 
middle  term  provided  the  equation  is  complete.  If  corresponding 
terms  have  like  signs,  the  middle  term  satisfies  the  condition  that 
terms  equally  distant  from  the  ends  are  equal,  and  the  equation 
is  of  the  first  kind ;  but  if  corresponding  terms  have  unlike  signs, 
the  middle  term  cannot  satisfy  the  condition  that  terms  equally 
distant  from  the  ends  have  unlike  signs  except  when  it  is  zero. 
In  this  case,  then,  the  middle  term  is  missing  and  f(x)  takes  the 
form 

f{x)  =  i(a;"  - 1)  +  sa;(aj"-2  —  1)  +  ra^(x''-*  -  1)  H . 

Since  n,  n  —  2,  n  — 4,  •••  are  even,  by  §  111,  1  and  2,  f(x)  is 
divisible  by  both  x  —  1  and  x  +  1.  Hence  1  and  —  1  are  roots  and 
the  equation  is  of  the  fourth  kind. 

It  should  be  observed  that  only  standard  reciprocal  equations 
have  an  odd  number  of  terms  in  the  first  member. 

702.    Let     pffl?"^  +2)iar''"-i  _| f-p^x™  ^ \-p^x  +Po  =  0       (1) 

represent  a  standard  reciprocal  equation. 

Grouping  corresponding  terms  and  dividing  by  «"•, 

Po(a^  +  a;-™)  +  p,{x^-'  +  a^-C"-^))  +  •  •  •  +  i>„_,(a^  -f  x-"^ 

+P^-i{x+x-')+p„  =  Q.  (2) 

By  the  binomial  formula, 

{x  +  x-'^f  =  a^  +  2  +x-'-  =  (a^  +  a;-^)  +  2, 

(a:  -f  x-y  =  ar'  +  3  a;  +  3  x-i  +  --c"'  =  (x"  +  x'^)  +  3  (a;  +  a;"^), 

{x  +  x-y  =  «*  +  4  a,-2  +  G  +  4  a;-2  +  x'*  =  (a;*  +  x'')  +  4(a^  +  x'")  +  6, 

and  so  on.     Transposing  and  substituting  y  for  x  +  x~'^, 

x"  -f-  .T-2  =  (;c  +  x-y  -2  =  f-1, 

x"  +  x'^  =  {x-\-  x-y  -  3(a;  +  x"^)  =  y^-3y, 

a^  +  X-*  =  (x  +  x-y  -  W  +  x-^)  -  6 

=  2^-%2- 2) -6  =  2/^-4/4-2, 


THEORY  OF  EQUATIONS  601 

and  so  on.  Continuing  this  process  until  the  value  of  a?"*  +  a?"*"  is 
found  in  terms  of  t/™  and  lower  powers  of  y,  and  substituting  all 
these  values  in  (2),  the  transformed  equation  after  reduction  takes 
the  form 

M™  +  qiy""^  ^ h  qm-\y  +  ^m  =  0.     Hence, 

Principle  9.  —  Every  standard  reciprocal  equation  may  he  trans- 
formed into  an  equation  of  half  its  degree. 

Other  reciprocal  equations  may  be  depressed  to  the  standard  form  by 
dividing  by  x  +  1,  x  —  1,  or  both  (§  699),  and  then  transformed  to  equations 
of  half  the  degree  of  the  depressed  equations. 

Examples 

1.  Solve  10 a;*- 77 a^  + 150 a;2- 77 a; +  10  =  0. 
Solution.  —  Grouping  the  terms  with  like  coefficients, 

10(x*  +  1)  _  77  x(x^  +  1)  +  150x2  =  0. 
Dividing  both  members  by  x^, 

Putting  in  the  quadratic  form  by  adding  and  subtracting  20, 
10(x2  +  2  +  l^-77^x  +  i]+  130  =  0. 

Let  X  +  -  =  y  ;  then,  10  j/2  -  77  y  +  130  =  0. 

X 

Factoring,  (2  y  -  5)  (5  y  -  26)  =  0. 

,.,^,  +  U5^,26. 
X      2         5 
Solving  these  two  equations  as  quadratics,  x  =  2,  ^,  5,  ^. 

The  given  equation  is  a  standard  reciprocal  equation. 

2.  Solve  10a^-67a;*-|-73ar'^  +  73a^-67a;  +  10  =  0. 

Solution.  —  Grouping  terms  with  like  coefficients,  it  is  evident  that  the 
first  member  is  divisible  by  x  +  1. 
Depressing  to  the  standard  form, 

10-67+    73+    73-67  +  101-1 

-  10  +    77  -  150  +  77-10 
10  -  77  +  150  -    77  +  10 
The  rest  of  the  solution  is  as  given  for  Ex,  1. 
.•.x  =  -l,  2,  i,  6,^. 


602  THEORY  OF  EQUATIONS 

3.  Solve  a:"  +  2 ic* - 4 »*  + 8 a;'' - 4 aj2-f  2a; +  1=0. 

Solution.  —  The  equation  Is  a  standard,  reciprocal  equation. 
Grouping  terms  with  like  coefficients  and  dividing  by  x^, 

(^+i)  +  2(..+  i)-4(.  +  l)+8=0. 

If  ic  +  -  =  2/,  as  in  §  702,  a;^  +  —  =  w2  -  2,  and  x^  +  j_  -  yS  _  3  „ 
X  x:^  'x? 

Substituting  these  values  and  reducing,  the  given  equation  is  transformed 
into  the  cubic  equation 

y8  +  2  j/2  _  7  2,  +  4  =  0, 

whose  roots,  by  trial,  are  found  to  be  1,  1,  —  4. 
Solving  the  quadratics  y  =  x  +  -  =  l,l,  or—  4, 

X 

a;=Kl+ v^3),  ^(l_V33),  ^(i  +  V^),  1(1 -V^),  -2+V3,  -2-^3. 
The  student  should  show  that  these  roots  are  reciprocals  in  pairs. 

4.  Solve  2a;^-15a;'»  +  37x3_37a^^l5a,_2==o. 

5.  Solve  12a;^  +  23x^-135ar''-135a;2^23a;  +  12  =  0. 

6.  Solve  a;«-2aT'-7a;*  +  7ar^  +  2a;-l  =  0. 

7.  Solve  a;«  +  a;^-14a;*  +  17a;3-14ar'  +  a;  +  l=0. 

8.  Transform a;*  +  a;^4-5a;^  +  4a;54-9a;*  +  4af'  +  5a:2  +  a;  +  l  =  0 
into  a  quadratic  equation. 

BINOMIAL  EQUATIONS 

703.  An  equation  of  the  form  a"  =  a,  or  a;"  —  a  =  0,  is  called  a 
Binomial  Equation. 

Thus,  x^  —  1  =  0  and  x^  +  1  =  0  are  binomial  equations. 

704.  The  n  nth.  roots  of  a  real  number. 

1.  The  n  nth  roots  of  unity  are  the  n  roots  of  the  equation 
iB»  -  1  =  0. 

By  Descartes'  rule,  if  n  is  odd,  this  equaition  has  one  real  root, 
+ 1,  and  n—1  imaginary  roots ;  if  n  is  even,  it  has  two  real  roots, 
-f- 1  and  —  1,  and  n  —  2  imaginary  roots. 


THEORY  OF  EQUATIONS  603 

Since  af  —  1  and  its  first  derived  function  nx^~^  have  no  com- 
mon factor,  af  —  1  has  no  multiple  roots. 

Hence,  the  n  nth  roots  of  unity  are  all  imaginary  except  one  when 
n  is  odd  or  two  when  n  is  even,  and  all  are  different. 

2.  The  n  nth.  roots  of  any  real  number  a  are  the  n  roots  of  the 
binomial  equation  af  =  a,  or  ic"  —  a  =  0. 

Let  any  nth  root  of  unity  be  represented  by  a  (the  Greek  letter 
alpha)  and  let  Va  be  the  principal  nth  root  of  a. 

Then,  (a  Va)»  =  a"  ( Va)"  =  1  .  a  =  a ; 

that  is,  the  n  nth  roots  of  a  are  found  by  multiplying  the  corre^ondr 
ing  nth  roots  of  unity  by  the  principal  nth  root  of  a. 

705.  Relation  between  the  n  nth  roots  of  unity. 
Let  a  be  any  imaginary  nth  root  of  unity. 
Then,  a"  =  1, 

and  if  m  is  a  positive  integer, 

(a")"*  =  1,  or  (tt")"  =  1 ; 
that  is,  a"  is  an  nth  root  of  unity. 

Hence,  any  power  of  an  imaginary  nth  root  of  unity  is  equal  to 
some  one  of  the  nth  roots  of  unity. 

706.  The  three  cube  roots  of  a  number. 

On  page  299  it  is  shown  that  the  three  cube  roots  of  1  are  1, 
\{—  1  +V— 3),  and  |-(— 1— V— 3),  the  roots  of  tiie  equation 
x^  -  1  =  0. 

The  student  may  verify  the  following  statements : 

The  sum  of  the  cube  roots  of  1  is  zero. 

The  two  imaginary  cube  roots  of  1  are  conjugates,  squares,  and 
reciprocals  of  each  other. 

The  cube  roots  of  1  are  equal  to  any  three  consecutive  powers  of 

i(-i+V^).  .  __ 

In  the  following  pages  |(— l+V—  3)  will  be  represented  by  w 
(the  Greek  letter  omega),  and  [i(  — 1  + V^)]^  or  i(-l  — V^^), 
by  o)''. 

Then,  the  cube  roots  of  1  are  1,  <o,  and  w^;  or  w,  wVand  w^;  or 
ii?,  <o',  and  to* ;  etc.     For  w^  =  1,  w*  =  w^  •  w  =  w,  etc. 


604  THEORY  OF  EQUATIONS 

Let  a  be  any  of  the  cube  roots  of  I.  Since  the  equation 
a?  —  a?  =  ()  gives  the  three  cube  roots  of  a^,  or  of  I,  and  since  this 
equation  is  derived  from  a;^  —  1  =  0  by  multiplying  the  roots  of 
the  latter  by  a,  the  cube  roots  of  I  are  a,  aw,  and  aur. 

Hence,  the  three  cube  roots  of  any  number  are  obtained  from  any 
one  of  them  by  multiplying  by  1,  w,  and  w^,  respectively. 
Thus,  the  three  cube  roots  of  64  are  4,  4  w,  and  4  w^. 

ALGEBRAIC  SOLUTIONS 

707.  So  far  the  theory  of  equations  has  been  applied  to  the 
solution  of  numerical  equations  and  has  been  found  sufficient  to 
determine,  exactly  or  approximately,  the  real  roots  of  numerical 
equations  of  any  degree.  In  the  following  pages  the  theory  of 
equations  will  be  applied  to  the  solution  of  certain  literal  equations. 

708.  The  general,  or  algebraic,  solution  of  a  literal  equation  con- 
sists in  finding  such  an  expression  involving  the  general  coeffi- 
cients as  will  represent  all  of  the  roots  at  the  same  time. 

Thus,  the  algebraic  solution  of  ax^  +  fta;  +  c  =  0  consists  in  finding  the 

expression  x  = — which  has  two  values,  inasmuch  as  6^  _  4  ac 

2  a 
has  two  square  roots. 

It  is  only  a  matter  of  common  agreement  that  these  two  values  shall  be 
denoted  by  ±  before  the  radical,  which  is  then  taken  to  mean  the  positive 
square  root  only. 

In  the  following  pages  algebraic  solutions  of  cubic  and  biquadf 
ratic  equations  will  be  given.  In  1825  the  mathematician  Abel 
demonstrated  the  impossibility  of  obtaining  algebraic  solutions  of 
equations  of  higher  degree  than  the  fourth. 

709.  Cardan's  solution  of  the  cubic  equation. 

By  §§  673,  676,  the  general  cubic  equation  may  be  transformed 
into  an  equation  of  the  form 

x^-\-ax-\-b  =  0.  (1) 

Since  any  number  can  be  separated  into  two  parts  having  a 
given  product,  it  is  permissible  to  suppose  that  each  root  of  (1) 
may  be  divided  into  two  parts  whose  product  is   —  \a. 

Let  the  parts  be  y  and  z. 


THEORY  OF  EQUATIONS  605 

Then,                                             x  =  y+z,  (2) 

and                                               3  2/2:  =  —  a.  (3) 
Substituting  the  values  of  x  and  a  in  (1), 

2/3  +  28  4-  &  =  0.  (4) 
Irom  (3),  z  =  —  ^.     Substituting  in  (4), 


32/ 


^-2^  +  ^  =  ^' 


a' 


or  y6^6^_^=0.  (5) 

Solving  (5)  as  a  quadratic  in  2/^, 


whence,  by  (4),  ^  = " I  ^  V^+f  •  (7) 


'27      4 
Taking  the  cube  roots  in  (6)  and  (7),  by  (2), 


The  lower  signs  are  omitteopbecause  they  give  the  same  value 
as  the  upper  signs,  y  and  z  being  interchangeable. 

This  is  commonly  called  Cardan^ s  formula  for  the  roots  of  the 
cubic  equation  a?  +  ax  -\-h  =  Q. 

Since  every  number  has  three  cube  roots,  if  the  indicated  cube 

roots  in  (8)  are  taken  in  all  possible  combinations,  nine  values  of 

X  result.     But  by  (3)  the  parts  of  x  must  be  so  taken  that  their 

product  is  —  \a,  a  rational  number.     Tliis  relation  between  the 

parts  of  each  root  limits  the  number  of  roots  to  three. 

b        fc^      &^ 
For  representing  the  three  cube  roots  of  —  7;+\/;7^  +  -7  by 

h,  ho},  and  Ato^,  and  the  three  cube  roots  of  the  conjugate  expression 
by  k,  k(i>,  and  kui^,  it  is  seen  that  h  •  k,  hw  •  kw%  and  ho)^  •  kto  are 
rational  products  while  each  of  the  six  remaining  products,  Ji  •  kw, 
h  •  kw^,  hu)  •  k,  ho) '  km,  h(o^  •  k,  and  hm^  -  kw^,  is  irrational. 


606  THEORY  OF  EQUATIONS 

Hence,  Cardan's  formula  is  to  be  applied  so  as  to  give  these 
roots 

X  =  h-\-  Jc,  hu)  +  ku)^,  hu)^  +  kw.  (9) 

Examples 

1.    Solve  a^-2ar'  +  48a;-96  =  0by  Cardan's  method. 

Solution.  —  Multiplying  the  roots  by  |,  Prin,  6, 

2/3  -  3  2/2  +  108  2/  -  324  =  0, 

in  which  y  =  ^x.     Decreasing  the  roots  of  this  equation  by  1  (§  676),  the 
transformed  equation  lacking  the  second  term  is 

03  +  105  z  -  218  =  0, 

in  which  2  =  2/  —  l=|a;  —  1. 

Substituting  105  for  a  and  —  218  for  b  in  Cardan's  formula, 


«  =  ^109  +  V54766  +  Vl09  -  a/54756  =  ^^^iS  +  V- 126 
=  7  —  5or7w  —  6w2or7co2_5w 

=  2  or  -  1+  6V^  or  -  1  -  6V^^. 

But  since  2!  =  fx-l,  x  =  |(s  +  1). 

.-.  x  =  2,  4  V^  3,  -  4>/^. 

Solve  by  Cardan's  method : 

2.  ar*  4- 24  x  +  56  =  0.  4.    27  ar'-117cc2+105aj+49=0. 

3.  8x'-20x^  +  Ux-'S  =  0.     5.    7  a^  +  Wx^ +  12  x  +  4:  =  0. 

710.    Discussion  of  the  roots  of  a  cubic  equation. 


the  three  roots  of  a^  +  ax  -\-b  =  0  are,  §  709,  (9),  h  +  k,  hw  +  ku,^ 
and  hoi^  +  k<a. 

52  ... 

1.    Since  —  is  always  positive,  if  a  is  positive  or  if  a  is  negative 

a^  W 

and  —  is  numerically  less  than  — ,  h  and  k  are  real  and  unequal. 

27  -^4 

Hence,  one  root  is  real  and  the  other  two  are  imaginary. 

Each  of  the  equations  x3  +  6x  +  2  =  0  and  x3_2x-4  =  0  has  one  real 
and  two  imaginary  roots. 


THEORY  OP  EQUATIONS  607 

2.  If  a  is  negative  and— ::r  is  numerically  equal  to  — ,  the  nuni- 

27  i 

bers  h  and  A;  are  equal  and  consequently  hu)  +  kuy^  —  hm^  +  ka) ; 
also,  /i  and  A;  are  real,  and  since  w  +  w^  =  —  1,  h  +  k,  h<a  +  fcw^, 
and  /io)^  +  fcw  are  real. 

Hence,  the  roots  are  real  and  two  of  them  are  equal. 

The  roots  of  x^  —  12  x  +  16  =  0  are  real  and  two  of  them  are  equal. 

3.  If 1 —    is  negative,  h?  and  T^  are  complex  numbers. 

Since  there  is  no  algebraic  method  of  finding  the  cube  root  of  a 
complex  number,  this  is  called  the  Irreducible  Case. 

In  this  case,  however,  the  roots  may  be  found  by  trigonometri- 
cal formulae,  and  it  may  be  proved  that  the  roots  are  real  and 
unequal.  In  the  irreducible  case,  then,  it  is  easy  to  find  the  roots 
by  trial  or  by  Horner's  method. 

By  Cardan's  formula  the  roots  ofx*  —  7x  +  6  =  0  are  given  by 


X  =  V-  3  +  V-  -\^  +  9  +  v  -  3  -  V-  -W-  +  9 

=  ^-  3  +  ^  y/^^i  +  a/ _  3  -  ify/^^, 

which  we  cannot  evaluate  by  algebra.     But  by  trial  the  roots  are  found  to  be 
1,  2,  and  —  3. 

711.   Descartes'  solution  of  the  biquadratic  equation. 

By  §§  673,  676,  the  general  biquadratic  equation  may  be  trans- 
formed into  an  equation  of  the  form 

X*  -f  qa?  +  ra;  -f  s  =  0.  (1) 

Assume  a^  -{-  q:x?  -\-  rx  -\-  s  =  {a?  +  Ax  +  B)(a^  —  Ax  -^  C),  (2) 

in  which  A,  B,  and  O  are  to  be  determined.     Expanding  (2), 

x!'  +  qa^  +  rx  +  s  =  x*  +  (- A""  +  B  +  C)x'  +  {AC  -  AB)x  +  BC, 

from  which,  §  590,  -A^  +  B+C=^q,AC-AB  =  r,BC==s.  (3) 

Eliminating  B  and  C  from  the  equations  in  (3), 

^«  -I-  2  qA^  -f-  (^2  -  4  s)A^  -7-2  =  0.  (4) 

Equation  (4)  is  a  cubic  in  A?.  Hence,  §  685,  Cor.  1,  A^  always 
has  at  least  one  positive  real  value,  and  when  this  has  been  found, 
the  values  of  B  and  C  may  be  found  by  means  of  (3).     Finally, 


608  THEORY  OF  EQUATIONS    . 

substituting  the  values  of  A,  B,  and  G  in  (2),  the  roots  of  the 
given  biquadratic  are  obtained  by  solving  the  quadratics 

x'  +  Ax  +  B  =  02in&x'-Ax+C  =  0. 

Equation  (4)  is  called  the  reducing  cubic  of  (1). 

Examples 

1.    Solve  the  equation  a;''-6a^+3a^  +  22a;-6  =  0. 

Solution.  —  Multiplying  the  roots  by  2  so  that  the  coefficient  of  the  third 
power  of  the  unknown  number  may  be  a  multiple  of  4, 

yi  _  12  2/3  +  12  2/2  +  176  2/  -  96  -  0,  in  which  ?/  =  2  x. 

Decreasing  the  roots  of  this  equation  by  J^^,  or  3, 

0*  -  42  22  ^.  32  2  +  297  =  0,  in  which  2  =  2  x  -  3, 

Since  this  equation  is  in  the  form  of  (1),  g  =  —  42,  r  =  32,  and  s  =  297. 
Substituting  these  values  in  the  reducing  cubic  (4), 

^6  -  84  ^*  +  576  ^2  _  1024  =  0. 
Simplifying  by  dividing  the  roots  by  4, 

Ai^  -  21  Ai^  +  36  ^1  -  16  =  0,  in  which  Ai  =  \  A^. 
By  trial,  vli  =  1  is  found  to  be  one  root  of  this  equation. 

.•.  \  J.2  =  1,  whence  A=  ^/^•\  =  2. 
Substituting  2  for  A,  -  42  for  g,  32  for  r,  and  297  for  s  in  (3), 

B+  C  =  -38,  C-5=16,  andBC=297. 
Solving,  B  =  -  27  and  C  =  -  11. 

2*  -  42  ^2  4-  32  z  +  297  =  {z^  +  Az  +  B){z^  -  Az  +  C) 

=  (22  +  2«-27)(22_2«-ll) 

Since  2  =  2  x  -  3,  =  16(x2  -  2  x  -  6)(x2  -  4  x  +  1). 

That  is,         X*  -  6  x3  +  3  x2  +  22  X  -  6  =  16Cx2  -  2  x  -  6)(x2  -  4  x  +  1). 
Equating  the  quadratic  factors  to  zero  and  solving, 

X  =  1  ±  V7,  2  ±  V3. 

Solve  the  following  equations : 

2.  x*-2x'-^x-'6  =  0. 

3.  a^-4ar'-8a;  +  35  =  0. 

4.  x""  -  12 :(? -^  ^5  a?  -  102  a;  +  124  =  0. 

5.  x^- 6x3 +  11  3^^-100;  + 2  =  0. 

6.  a;*-2x3-15ar^  +  16a;  +  14  =  0. 


Il 


I 


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